LeetCode-Sum of Square Numbers

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Description：
Given a non-negative integer c, your task is to decide whether there’re two integers a and b such that a2 + b2 = c.

Example 1:

Input: 5
Output: True
Explanation: 1 * 1 + 2 * 2 = 5

Example 2:

Input: 3
Output: False

题意：给定一个非负数 $c$，计算是否存在整数 $a$和 $b$，使得 $a^2 + b^2 = c$；

解法一（超时）：最简单的办法就是使用暴力求解， $a$和 $b$的取值区间在 $[0, c^{1/2}]$；最终肯定会超时的；

Java

class Solution {
    public boolean judgeSquareSum(int c) {
        for (long a = 0; a * a <= c; a++) {
            for (long b = 0; b * b <= c; b++) {
                if (a * a + b * b == c) {
                    return true;
                }
            }
        }
        return false;
    }
}

解法二：对公式变形得到 $b^2 = c - a^2$，因此我们只需要在区间 $[0, c^{1/2}]$内寻找是否由满足条件的 $b$即可；这里我们利用二分查找来实现快速的查找；

Java

class Solution {
    public boolean judgeSquareSum(int c) {
        for (long a = 0; a * a <= c; a++) {
            long b = c - a * a;
            if (find(0, b, b)) {
                return true;
            }
        }
        return false;
    }
    
    private boolean find(long low, long high, long target) {
        if (low > high) {
            return false;
        }
        while (low <= high) {
            long mid = low + (high - low) / 2;
            if (mid * mid == target) {
                return true;
            } else if (mid * mid > target) {
                high = mid - 1;
            } else {
                low = mid + 1;
            }
        }
        return false;
    }
}