Leetcode 4

Array Easy

1. 268. Missing Number

　　先对数组求和，用 0 ~ n本该有的和减去当前sum得到缺失的数字。

class Solution {
    public int missingNumber(int[] nums) {
        int sum = 0;
        for ( int i = 0 ; i < nums.length ; i++){
            sum += nums[i];
        }   
        return ((nums.length) * (nums.length + 1 )/2) - sum;
    }
}

2. 169. Majority Element

　　利用HashMap键值对记录次数，并用最大次数大于n/2来判定满足条件

class Solution {
    public int majorityElement(int[] nums) {
        int len = nums.length;
        if( len == 1)
            return nums[0];
        
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for( int i = 0; i < nums.length ; i++){
            if(!map.containsKey(nums[i])){
                map.put(nums[i],1);
            }else{
                int times = map.get(nums[i]);
                map.put(nums[i], times+1);
                if(times+1 > len/2)
                    return nums[i];
            }
        }
        return 0;
    }
}

3. 189. Rotate Array 

　　先把整个数组翻转一下, 再把前k个数字翻转一下，再把后n - k个数字翻转一下。翻转的逻辑可以梳理一下

class Solution {
    public void rotate(int[] nums, int k) {
        int len = nums.length;
        k = k % len;
        reverse(nums, 0, len-1);
        reverse(nums, 0, k -1);
        reverse(nums, k, len-1);
    }
    
    public static void reverse(int[] nums, int l, int r){
        while( l < r){
            int temp = nums[l];
            nums[l] = nums[r];
            nums[r] = temp;
            l++;
            r--;
        }
    }
}