问题描述
https://leetcode.com/problems/sqrtx/#/description
Implement int sqrt(int x)
.
Compute and return the square root of x
.
算法
使用折半查找即可,但要注意整型溢出
代码
public int mySqrt(int x) {
int left = 1, right = x;
while(left<=right) {
int mid = left + (right-left)/2;// 不能使用(right+left),因为有可能整数溢出
if(mid <= x/mid) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return left-1;
}