leetcode 448

448. Find All Numbers Disappeared in an Array

Input:
[4,3,2,7,8,2,3,1]

Output:
[5,6]

思路：把数组的内容和index进行一一对应映射，映射规则是取反，由此可知，重复出现两次的数字会变为正，出现一次的为负。

需要注意的是，如果不能增加额外空间的话，要在本数组上面进行映射，这时候就需要内容-1=index，因为index是从0开始，内容从1开始

class Solution {
public:
    vector<int> findDisappearedNumbers(vector<int>& nums) {
        int len = nums.size();
        for(int i=0; i<len; i++) {
            int m = abs(nums[i])-1; // index start from 0
            nums[m] = nums[m]>0 ? -nums[m] : nums[m];
        }
        vector<int> res;
        for(int i = 0; i<len; i++) {
            if(nums[i] > 0) res.push_back(i+1);
        }
        return res;
    }
};

455. Assign Cookies

Input: [1,2], [1,2,3]

Output: 2

Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
You have 3 cookies and their sizes are big enough to gratify all of the children, 
You need to output 2.

思路：先对小孩和饼干进行排序，逐个对应匹配。若当前小孩符合，则下一个小孩和下一个饼干；否则下一个饼干匹配当前小孩。

class Solution {
public:
      int findContentChildren(vector<int>& g, vector<int>& s) {
        sort(g.begin(),g.end());
        sort(s.begin(),s.end());
        int i = 0, j=0;
        while(i<g.size() && j<s.size()){
            if(s[j]>=g[i])
                i++; // when the child get the cookie, foward child by 1
            j++;
        }
        return i;
    }
};

 461. Hamming Distance

Input: x = 1, y = 4

Output: 2

Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑

The above arrows point to positions where the corresponding bits are different.

class Solution {
public:
    int hammingDistance(int x, int y) {
        int dist = 0, n = x ^ y; //按位异或
        while (n) {
            ++dist;
            n &= n - 1; //效果循环右移，左补零
        }
        return dist;
    }
};

459. Repeated Substring Pattern

Input: "abab"
Output: True
Explanation: It's the substring "ab" twice.

思路：

法一：从str的一半开始，用

class Solution {
public:
    bool repeatedSubstringPattern(string str) {
        string nextStr = str;
        int len = str.length();
        if(len < 1) return false;
        for(int i = 1; i <= len / 2; i++){
            if(len % i == 0){  //若能整除，表示源字符串可以划分成正数个子字符串
                nextStr = leftShift(str, i); //将源字符串左移一个子字符串，若左移后和源字符串相等，则判断为符合；这里也可以使用一个子字符串乘以个数得到的串和源字串比较
                if(nextStr == str) return true;
            }
        }
        return false;
    }
    
    string leftShift(string &str, int l){
        string ret = str.substr(l);
        ret += str.substr(0, l);
        return ret;
    }
};

法二：

将源字符串变成两倍，然后去掉首尾字符，剩下的串里面寻找源字串，若存在则判符合。

内部思想，如果源字符串不存在重复子字符串，则两倍后的字符串去掉前后字母后，里面必定不存在源字符串；否则必定存在源字符串。

bool repeatedSubstringPattern(string str) 
    {
        return (str + str).substr(1, str.size() * 2 - 2).find(str)!=-1;
    }

476. Number Complement

Input: 5
Output: 2
Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.

//使用mask来统计num二进制的位数
class Solution {
public:
    int findComplement(int num) {
        unsigned mask = ~0;
        while (num & mask) mask <<= 1;
        return ~mask & ~num;
    }
};
For example,

num          = 00000101
mask         = 11111000
~mask & ~num = 00000010

//使用i来统计num二进制的位数
class Solution(object):
    def findComplement(self, num):
        i = 1
        while i <= num:
            i = i << 1
        return (i - 1) ^ num