367. Valid Perfect Square 发表于 2022-02-03 Binary Search1234567891011121314151617class Solution { public boolean isPerfectSquare(int num) { int left = 1, right = num; int ans = 0; while (left <= right) { int mid = (left + right) >>> 1; if (mid <= num / mid) { ans = mid; left = mid + 1; } else { right = mid - 1; } } return ans * ans == num; }} Math1234567891011class Solution { public boolean isPerfectSquare(int num) { int x = 1; while (num > 0) { num -= x; x += 2; } return num == 0; }} Reference367. Valid Perfect Square