C++ leecode

两种解法：左闭右闭，左闭右开。重点区间定义。
解法一

class Solution {
public:
    int search(vector& nums, int target) {
        int len = nums.size();
        int a = 0;
        int b = len-1;
        
        while (a<=b)  //左闭右闭
        {
            int m = a + ((b-a) / 2); //防止溢出
            if (nums[m] > target)
                b = m-1;
            else if (nums[m] < target)
                a = m+1;
            else
                return m;
        }
        return -1;
    }
};

解法二：

class Solution {
public:
    int search(vector& nums, int target) {
        int a = 0;
        int b = nums.size();
        
        while (a target)
                b = m; //右开无需加一
            else if (nums[m] < target)
                a = m+1;
            else
                return m;
        }
        return -1;
    }
};

##977

非递减数列，其平方最大值一定在左右两边上。

class Solution {
public:
    vector sortedSquares(vector& nums) {
        int i = 0;
        int j = nums.size()-1;
        int k = j;
        vector result(j+1, 0);
        while (i <= j){
            int ti = nums[i]*nums[i];
            int tj = nums[j]*nums[j];
            if (ti < tj){
                result[k--] = tj;
                j--;
            }
            else {
                result[k--] = ti;
                i++;
            }
        }
        
        return result;
    }
};