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72.mySqrt.cpp
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executable file
·73 lines (69 loc) · 1.61 KB
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/*
@filename 72.mySqrt.cpp
@author caonan
@date 2022-05-05 15:06:53
@reference 剑指offer专项
@url https://leetcode-cn.com/problems/jJ0w9p
@brief 给定一个非负整数 x ，计算并返回 x 的平方根，即实现 int sqrt(int x) 函数。
正数的平方根有两个，只输出其中的正数平方根。
如果平方根不是整数，输出只保留整数的部分，小数部分将被舍去。
*/
#include <assert.h>
#include <stdio.h>
#include <algorithm>
#include <bitset>
#include <deque>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#include <vector>
using namespace std;
class Solution {
public:
int mySqrt(int x) {
int l = 1;
int r = x;
while (l <= r) {
int mid = l + (r - l) / 2;
if (mid <= x / mid && mid + 1 > x / (mid + 1)) {
return mid;
}
if (mid < x / mid) {
l = mid + 1;
} else {
r = mid - 1;
}
}
return 0;
}
};
class Solution2 {
public:
int mySqrt(int x) {
int left = 0;
int right = x;
int ans = -1;
while (left <= right) {
int t = left + (right - left) / 2;
// 这里要注意一个整形溢出的问题,如果t是int，那么t*t可能会溢出，即使接收的变量是long long
// ，但是t*t还是int，所以要强制转换
if ((long long)t * t <= x) {
ans = t;
left = t + 1;
} else {
right = t - 1;
}
}
return ans;
}
};
int main() {
Solution s;
cout << "ret is " << s.mySqrt(1) << endl;
return 0;
}