LeetCode - Valid Perfect Square

brute force从1算到n^2(>num)

解法一 缩小1-n^2的范围

找到前后区间

class Solution {
public:
    bool isPerfectSquare(int num) {
        if(num==1) return true;
        long cur=num;
        while((cur/2)*(cur/2)>num){
            cur = cur/2;
        }
        int h=cur;
        for(int i=cur/2;i<cur;i++){
            if(i*i==num) return true;
        }
        return false;
    }
};

边遍历边缩小区间

class Solution {
public:
    bool isPerfectSquare(int num) {
        for (int i = 1; i <= num / i; ++i) {
            if (i * i == num) return true;
        }
        return false;
    }
}; 

解法二 Binary Search

class Solution {
public:
    bool isPerfectSquare(int num) {
        if(num==1) return true;
        long l=0, r=num/2+1;
        while(l<r){
            long mid = l+(r-l)/2;
            long cur = mid*mid;
            if(cur==num) return true;
            else if(num>cur) l=1+mid;
            else r=mid;
        }
        return false;
    }
};

解法三 Math: 平方是连续的奇数和

1 = 1
4 = 1 + 3
9 = 1 + 3 + 5
16 = 1 + 3 + 5 + 7
25 = 1 + 3 + 5 + 7 + 9
36 = 1 + 3 + 5 + 7 + 9 + 11
....
1+3+...+(2n-1) = (2n-1 + 1)n/2 = n*n

class Solution {
public:
    bool isPerfectSquare(int num) {
        int i = 1;
        while (num > 0) {
            num -= i;
            i += 2;
        }
        return num == 0;
    }
};

class Solution {
public:
    bool isPerfectSquare(int num) {
        long x = num;
        while (x * x > num) {
            x = (x + num / x) / 2;
        }
        return x * x == num;
    }
};

O(1)

Reference: