69. Sqrt(x)(二分查找)

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

class Solution {
public:
    int mySqrt(int x) {
        if (x<=1) return x;
        int low = 0,high = x;
        while(low < high) {
            int mid = low + (high - low)/2;
            int aa = x/mid;
            if (aa<mid) {
                high = mid;
            } else if (aa > mid ) {
                low  = mid + 1;
            } else {
                return mid;
            }
        }
        // 开区间，打个补丁
        if (low>x/low) return low-1;
        return low;
    }
};

 

class Solution {
public:
    int mySqrt(int x) {
        if (x <=1) return x;
        int low = 0;
        int high = x ;
        while(low <= high) {
            int mid = low + (high-low)/2;
            if (x/mid > mid) {
                low = mid + 1;
            } else if(x/mid < mid) {
                high = mid - 1;
            } else {
                return mid;
            }
        }
        // 正常二分法，如果没找到，返回的是大于target 的位置。此题需要返回小于target的位置。8 的平方根是 2.82842..., 由于返回类型是整数，小数部分将被舍去。
        return low - 1;
    }
};