Lintcode697-判断是否为平方数之和

来源：互联网发布：navicat 备份数据库编辑：程序博客网时间：2024/06/06 19:44

哈哈，这题网上竟然没有找到答案，于是自己copy到VS里面调试了几次终于AC了，不保证是最优的算法，时间复杂度应该是O(nlogn)，空间复杂度O(1)（循环里面嵌套了一个二分查找），万一有人需要呢，分享一下~

class Solution {public:    /*     * @param : the given number     * @return: whether whether there're two integers     */  bool isSquare(int n) {if (n < 0) {return false;}if (n == 0) return true;int low = 1;int high = n;int mid = (low + high) >> 1;while (low <= high) {int temp = n / mid;if (mid == temp && n == mid * temp) {return true;}if (temp < mid) {high = mid - 1;}if (temp >= mid) {low = mid + 1;}mid = (low + high) >> 1;}return false; // fix me!  }bool checkSumOfSquareNumbers(int num) {// write your code hereif (num<0) return false;if (isSquare(num)) return true;for (int i = 1; i<=sqrt(num); i++){int target = (num - i * i);if (!isSquare(target)) continue;int j = sqrt(num)+1;int mid = 0;while (i <= j){mid = (i + j) / 2;if (sqrt(target)>mid)i = mid + 1;else if (sqrt(target)<mid)j = mid - 1;elsereturn true;}if (mid != sqrt(target)) return false;elsereturn true;}}};

终于会用二分查找了~

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