leetcode 69 - Sqrt(x)

leetcode题目汇集 —  Sqrt(x)

leetcode 69 - Sqrt(x)

题目要求：

int sqrt(int x)

注意：程序中需要进行乘法计算，为避免数值逸出，中间数值应设置为long

解法1：从1开始遍历，直到root的平方大于x，此时返回结果。时间复杂度为o(n)

public class Solution {
public int mySqrt(int x) {
if (x < 0) {
return -1;
}
if (x == 0) {
return 0;
}
int root = 1;
long result = root * root;
while (result > 0 && result <= x) {
root++;
result = root * root;
}
return root - 1;
}
}

解法2：二分法，时间复杂度为o(log n)

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