题目描述

• 本质：找满足 k * k <= x 的最大 k
  

思路 & 代码

• 二分，对满足 k * k <= x 的最大 k进行查找。
• 注意极端数据，用 long
• 暴力法先来一个

class Solution {public int mySqrt(int x) {int ans = -1;for(int i = 0; i <= x; i++){if((long)i * i <= x){ans = i;}else{break;}}return ans;}
}

• 二分

class Solution {public int mySqrt(int x) {int ans = -1;int left = 0, right = x;// 二分法：查找最大的，满足 mid * mid <= x 的值为answhile(left <= right){int mid = (left + right) / 2;// long：极端数据if((long)mid * mid <= x){ans = mid;left = mid + 1;}else{right = mid - 1;}}return ans;}
}

更新版

• 只能说这道题真的挺恶心的…本质还是二分，小心边界。

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