Math

4月 21日

1  7 Reverse Integer  先算新结果，与旧结果比较，如果不等 返回0

    public int reverse(int x) {
        int result = 0;
        while (x != 0)
        {
            int tail = x % 10;
            int newR = result * 10 + tail;
            if ((newR - tail)/10!= result)
            {
                return 0;
            }
            result = newR;
            x /= 10;
        }
        return result;
    }

2  8 String to Integer  去空格，找符号 转大小

    public int myAtoi(String str) {
        int index = 0, sign = 1, sum = 0;
        if (str == null || str.length() == 0) return 0;
        while (index < str.length() && str.charAt(index) == ' ')
        {
            index++;
        }
        if (str.charAt(index) == '+' || str.charAt(index) == '-')
        {
            sign = str.charAt(index) == '+' ? 1:-1;
            index++;
        }
        while (index < str.length())
        {
            if (!Character.isDigit(str.charAt(index))) return sign * sum;
            int digit = str.charAt(index) - '0';
            if (Integer.MAX_VALUE / 10 < sum || Integer.MAX_VALUE / 10 == sum &&  Integer.MAX_VALUE % 10 < digit)
            {
                return sign == 1? Integer.MAX_VALUE : Integer.MIN_VALUE;
            }
            sum = sum * 10 + digit;
            index++;
        }
        return sign * sum;
    }

3  9 Palindrome Number  检测小于0 或 10的倍数  当rev < x

    public boolean isPalindrome(int x) {
        if (x < 0 || x != 0&&x % 10 == 0) return false;
        int rev = 0;
        while (x > rev)
        {
            rev = rev * 10 + x%10;
            x/=10;
        }
        return (x==rev || x==rev/10);
    }

4 12 Integer to Roman   写出各个位的10位表示 相加

    public String intToRoman(int num) {
    String M[] = {"", "M", "MM", "MMM"};
    String C[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
    String X[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
    String I[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
    return M[num/1000] + C[(num%1000)/100] + X[(num%100)/10] + I[num%10];
        
    }

 4 月 22 日

5 13 Roman to Integer   比较俩个相邻字符大小

    public int romanToInt(String s) {
        int sum = 0;
        for (int i = 0; i < s.length() - 1; i++)
        {
            int i1 = get(s.charAt(i)), i2 = get(s.charAt(i + 1));
            if (i1 < i2)
            {
                sum -= i1;
            }else
            {
                sum += i1;
            }
        }
        return sum + get(s.charAt(s.length()-1));
    }
    public int get(char c)
    {
        switch(c)
        {
            case 'M':
                return 1000;
            case 'D':
                return 500;
            case 'C':
                return 100;
            case 'L':
                return 50;
            case 'X' :
                return 10;
            case 'V':
                return 5;
            case 'I':
                return 1;
            default:
            return 0;
        }
    }

6 29 Divide Two Integer  除数 <= 被除数 >> 1  找到digit 和除数界限 比较除数和被除数和最小值之间关系

    public int divide(int dividend, int divisor) {
         if (divisor == 0) return Integer.MAX_VALUE;
         boolean isN = (dividend ^ divisor) >>> 31 == 1;
         int res = 0;
         if (dividend == Integer.MIN_VALUE)
         {
             dividend += Math.abs(divisor);
             if (divisor == -1)
             {
                 return Integer.MAX_VALUE;
             }
             res++;
         }
         if (divisor == Integer.MIN_VALUE) return res;
         dividend = Math.abs(dividend);
         divisor = Math.abs(divisor);
         int digit = 0;
         while (divisor <= (dividend >> 1))
         {
             digit++;
             divisor <<= 1;
         }
         while(digit >= 0)
         {
             if (dividend >= divisor)
             {
                 res += 1 << digit;
                 dividend -= divisor;
             }
             digit--;
             divisor >>= 1;
         }
         return isN? -res:res;
    }

7 43 Multiply Strings  i+j, i+j+1  按照乘法规律  从后往前

    public String multiply(String num1, String num2) {
        int m = num1.length(), n = num2.length();
        int[] res = new int[m + n];
        for (int i = m - 1; i >= 0; i--)
        {
            for (int j = n - 1; j >= 0; j--)
            {
                int num = (num1.charAt(i) - '0')*(num2.charAt(j) - '0');
                int p1 = i + j, p2 = i + j + 1;
                num += res[p2];
                res[p1] += num / 10;
                res[p2] = num % 10;
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int i : res)
        {
            if (!(sb.length() == 0 && i == 0))
                sb.append(i);
        }
        return sb.length() == 0? "0" :sb.toString();
    }

8 50 Pow(x,n)  n >0 x平方

    public double myPow(double x, int n) {
        if (n == 0) return 1.0;
        double res = 1.0;
        if (n < 0)
        {
            if (x >= 1/Double.MAX_VALUE || x <= -1/Double.MAX_VALUE)
            {
                x = 1/x;
            }else
            {
                return Double.MAX_VALUE; 
            }
            if (n == Integer.MIN_VALUE)
            {
                res *= x;
                n++;
            }
        }
        n = Math.abs(n);
        boolean isN = false;
        if (n % 2 == 1 && x < 0) isN = true;
        x = Math.abs(x);
        while (n > 0)
        {
            if ((n & 1) == 1)
            {
                if (Double.MAX_VALUE / x < res)
                {
                    return Double.MAX_VALUE;
                }
                res *= x;
            }
            x *= x;
            n >>= 1;
        }
        return isN ? -res: res;
    }

 4月 23 日

9 60  Permutation Sequence   先减1，除（n-1）找索引，迭代

    public String getPermutation(int n, int k) {
        if (n < 0) return "";
        k--;
        List<Integer> num = new ArrayList<>();
        StringBuilder res = new StringBuilder();
        for (int i = 1; i <= n; i++)
        {
            num.add(i);
        }
        int count = n - 1, f = 1;
        for (int i = 2; i < n; i++)
        {
            f *= i;
        }
        while (count >= 0)
        {
            int index = k / f;
            k = k % f;
            res.append(num.get(index));
            num.remove(index);
            if (count > 0)
            {
                f /= count;
            }
            count --;
        }
        return res.toString();
    }

 10  65 Valid Number  四个标记数 e .  num在e后   几个判段

    public boolean isNumber(String s) {
        s = s.trim();
        boolean numsee = false;
        boolean pointsee = false;
        boolean esee = false;
        boolean numaftere = true;
        for (int i = 0; i < s.length(); i++)
        {
            char c = s.charAt(i);
            if (c >= '0' && c <= '9')
            {
                numsee = true;
                numaftere = true;
            }
            else if (c == '.')
            {
                if (pointsee || esee)
                {
                    return false;
                }
                pointsee = true;
            }
            else if (c == 'e')
            {
                if (esee || !numsee)
                {
                    return false;
                }
                esee = true;
                numaftere = false;
            }
            else if (c == '+' || c == '-')
            {
                if (i != 0 && s.charAt(i - 1)!= 'e')
                {
                    return false;
                }
            }
            else
            {
                return false;
            }
        }
        return numaftere && numsee;
    }

11 Add Binary    从后计算 反转

    public String addBinary(String a, String b) {
        int i = a.length() - 1, j = b.length() - 1, carry = 0;
        StringBuilder sb = new StringBuilder();
        while (i >= 0 || j >= 0)
        {
            int sum = carry;
            if (i >= 0) sum += (a.charAt(i--) - '0');
            if (j >= 0) sum += (b.charAt(j--) - '0');
            sb.append(sum%2);
            carry = sum/2;
        }
        if (carry != 0) sb.append(carry);
        return sb.reverse().toString();
    }

12  Sprt(x)  二分查找

    public int mySqrt(int x) {
        if (x < 0) return -1;
        if (x == 0) return 0;
        int l = 1, r = x / 2 + 1;
        while (l <= r)
        {
            int m = (r - l) / 2 + l;
            if (m <= x/m && (m + 1) > x /(m + 1))
            {
                return m;
            }
            else if (m > x / m)
            {
                r = m - 1;
            }
            else
            {
                l = m + 1;
            }
        }
        return 0;
    }

 4 月 24 日

13  166 Frction to Recurring Decimal   按除法操作 维护map 和 stringbuilder

    public String fractionToDecimal(int numerator, int denominator) {
        if (numerator == 0) return "0";
        StringBuilder sb = new StringBuilder();
        sb.append(((numerator > 0) ^ (denominator > 0)) ? "-":"");
        long num = Math.abs((long)numerator);
        long den = Math.abs((long)denominator);
        sb.append(num/den);
        num %= den;
        if (num == 0) return sb.toString();
        sb.append(".");
        HashMap<Long, Integer> res = new HashMap<>();
        res.put(num, sb.length());
        while (num != 0)
        {
            num *= 10;
            sb.append(num / den);
            num %= den;
            if (res.containsKey(num))
            {
                sb.insert(res.get(num), "(");
                sb.append(")");
                break;
            }
            res.put(num, sb.length());
        }
        return sb.toString();
    }

14 168 Excel Sheet Column Title   n--  对26取余  并+A

    public String convertToTitle(int n) {
        return n == 0? "" : convertToTitle(--n/26) + (char)('A'+(n%26));
    }

15 171 Excel Sheet Column Number  res * 26 + 余

    public int titleToNumber(String s) {
        int res = 0;
        for (int i = 0; i < s.length(); res = res * 26 + (int)(s.charAt(i) - 'A'+1), i++);
        return res;
    }

16 172 Factorial Trailing Zerses  除5

    public int trailingZeroes(int n) {
         return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
    }

17 202 Happy Number

    public boolean isHappy(int n) {
        int fast = n, slow = n;
        do {
            fast = cal(fast);
            fast = cal(fast);
            slow = cal(slow);
            if (fast == 1) return true;
        }while (fast != slow);
        return false;
        
    }
    public int cal(int n)
    {
        int res = 0;
        while (n != 0)
        {
            res += (n % 10) * (n % 10);
            n/= 10;
        }
        return res;
    }

 4 月 25 号

18  204 Count Primes   boolean数组 ，遍历 

    public int countPrimes(int n) {
        boolean[] isP = new boolean[n];
        int count = 0;
        for (int i = 2; i < n; i++)
        {
            if (!isP[i])
            {
                count ++;
                for (int j = 2; j * i < n; j++)
                {
                    isP[j * i] = true;
                }
            }
        }
        return count;
    }

19 223  Rectangle Area  两大面积减小面积

    public int computeArea(int A, int B, int C, int D, int E, int F, int G, int H) {
        int left = Math.max(A,E), right = Math.max(left, Math.min(C,G));
        int bottom = Math.max(B,F), top = Math.max(bottom, Math.min(D,H));
        return (C-A)*(D-B) - (right - left)*(top - bottom) + (G-E)*(H-F);
    }

20 224 Basic Calculator  用一个栈维护，每遇到（数放入栈中

    public int calculate(String s) {
        int res = 0, sign = 1;
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < s.length(); i++)
        {
            char c = s.charAt(i);
            if (Character.isDigit(c))
            {
                int sum = c - '0';
                while (i + 1 < s.length()&& Character.isDigit(s.charAt(i + 1)))
                {
                    sum = sum * 10 + (s.charAt(i + 1) - '0');
                    i++;
                }
                res += sum * sign;
            }
            else if(c == '+')
            {
                sign = 1;
            }
            else if(c == '-')
            {
                sign = -1;
            }
            else if(c == '(')
            {
                stack.push(res);
                stack.push(sign);
                res = 0;
                sign = 1;
            }else if (c == ')')
            {
                res = res * stack.pop()+stack.pop();
            }
        }
        return res;
    }

21 Power of Two   

    public boolean isPowerOfTwo(int n) {
        return ((n &(n - 1))==0 && n > 0);
    }

 4月26 号

22 233  Number of Digit One   公式

    public int countDigitOne(int n) {
    int ones = 0;
    for (long m = 1; m <= n; m *= 10)
        ones += (n/m + 8) / 10 * m + (n/m % 10 == 1 ? n%m + 1 : 0);
    return ones;
    }

23 246  Strobogrammtic number  把数放入map，比较左右

    public boolean isStrobogrammatic(String num) {
        HashMap<Character, Character> map = new HashMap<>();
        map.put('0','0');
        map.put('1','1');
        map.put('6','9');
        map.put('8','8');
        map.put('9','6');
        int l = 0, r = num.length() - 1;
        while (l <= r)
        {
            if (!map.containsKey(num.charAt(l)) || map.get(num.charAt(l))!=num.charAt(r))
            {
                return false;
            }
            l++;r--;
        }
        return true;
    }

24 247 Strobogrammtic number II  递归

public class Solution {
    public List<String> helper(int n, int m) {
        if (n == 0) return new ArrayList<String>(Arrays.asList(""));
        if (n == 1) return new ArrayList<String>(Arrays.asList("0", "1", "8"));
        
        List<String> list = helper(n - 2, m);
        List<String> result = new ArrayList<String>();
        for (int i = 0; i < list.size(); i++) {
            String s = list.get(i);
            
            if (n != m) result.add("0" + s + "0");
            result.add("1" + s + "1");
            result.add("8" + s + "8");
            result.add("6" + s + "9");
            result.add("9" + s + "6");
        }
        
        return result;
    }
    
    public List<String> findStrobogrammatic(int n) {
        return helper(n, n);
    }
}

25 248 Strobogrammtic number III

    public int strobogrammaticInRange(String low, String high) {
        int count = 0;
        List<String> res = new ArrayList<>();
        for (int i = low.length(); i <= high.length(); i++)
        {
            res.addAll(help(i,i));
        }
        for (String str : res)
        {
            if (str.length() == low.length() && str.compareTo(low) < 0 || str.length() == high.length() && str.compareTo(high) > 0)
            {
                continue;
            }
            count++;
        }
        return count;
    }
    private List<String> help(int m, int n) {
        // TODO Auto-generated method stub
        if (m == 0) return new ArrayList<String>(Arrays.asList(""));
        if (m == 1) return new ArrayList<String>(Arrays.asList("0", "1", "8"));
        List<String> list = help(n - 2, n);
        List<String> res = new ArrayList<>();
        for (int i = 0; i < list.size(); i++)
        {
            String s = list.get(i);
            if (m != n) res.add("0" + s + "0");
            res.add("1" + s + "1");
            res.add("6" + s + "9");
            res.add("8" + s + "8");
            res.add("9" + s + "6");            
        }
        return res;
    }