Leetcode 69. Sqrt(x)

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本文解析LeetCode第69题，实现整数平方根的计算，采用二分查找法在[1,x]区间内搜索，时间复杂度为O(logn)，适用于非负整数输入，通过实例演示算法过程。

Leetcode 69. Sqrt(x)

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.

题目大意：
给定一个数x，返回其平方根的整数部分。

解题思路：
答案一定来自于[1, x]区间内，区间有序则可以利用二分搜索压缩搜索空间。时间复杂度O(logn).

代码：

class Solution {
public:
    int mySqrt(int x) {
        if(x <= 1)
            return x;
        int l = 1, r = x - 1;
        while(l <= r)
        {
            int mid = (r - l) / 2 + l;
            if(int(x / mid) == mid)
                return mid;
            else if(int(x / mid) > mid)
                l = mid + 1;
            else
                r = mid - 1;
        }
        return r;
    }
};