【leetcode】69. Sqrt(x)

题目：
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.

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思路：
牛顿迭代法，如下图。

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代码实现：

class Solution {
public:
    int mySqrt(int x) {
        if (x == 0){
            return 0;
        }
        
        double ans = 1;
        double pre = 0;
        
        while (abs(ans-pre) >= 1.0){
            pre = ans;
            ans = (ans + x / ans) / 2;
        }
        
        return int(ans);
    }
};

discuss:
不同区间对应不同的处理策略。

class Solution {
public:
    int mySqrt(int x) {
        if (x == 0){
            return 0;
        }   
        
        int left = 1, right = INT_MAX;
        
        while (true){
            int mid = left + (right - left) / 2;
            if (mid > x / mid){ // x < mid^2
                right = mid - 1;
            }else if (mid + 1 > x / (mid + 1)){ // x < (mid+1)^2
                return mid;
            }else{
                left = mid + 1;
            }
        }
        
    }
};

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参考：https://www.cnblogs.com/grandyang/p/4346413.html