1: intsingleNumber(int A[], int n) { 2: int left = A[0]; 3:for(int i =1;i< n;i++) 4: { 5: left = left ^ A[i]; 6: } 7: return left; 8: }
于是利用交换律可以将数组假想成相同元素全部相邻,于是将所有元素依次做异或操作,相同元素异或为0,最终剩下的元素就为Single Number。时间复杂度O(n),空间复杂度O(1) 1classSolution {2public:3intsingleNumber(intA[],intn) {45//异或6intelem =0;7for(inti =0; i < n ; i++) {8elem = elem ^A[...
挣扎了一段时间,就去讨论区看解答(https://leetcode.com/problems/strong-password-checker/discuss/91003/O(n%29-java-solution-by-analyzing-changes-allowed-to-fix-each-problem)去了: 代码语言:javascript 代码运行次数:0 运行 复制 class Solution { public int strongPasswordChecker(String s) { int res...
(,, ): HDU 1542 Atlantis update: query: HDU 1828 Picture update: query: Title Solution Difficulty Time Space O(n log 218. The Skyline Problem Go Hard O(n) n) 307. Range Sum Query - Go Hard O(1) O(n) Mutable 315. Count of Smaller O(n log Go Hard O(n) Numbers After Self ...
classSolution:defisBalanced(self,root):""" 判断一个树是否为平衡二叉树 当check函数的发挥值不等于-1时返回true,等于-1是返回false:param root:TreeNode:return:bool"""returnself.check(root)!=-1defcheck(self,root):"""检查结点:param root:TreeNode:return:int""" ...
Scalability: A large number of users. 4 and 5 step will go in loop till we get a satisfactory answer Current Scenario Use cases for this problem. Parking can be single-level or multilevel. Types of vehicles that can be parked, separate spaces for each type of vehicle. ...
136Single NumberEasySolution.java 139Word BreakMediumSolution.java 146LRU CacheMediumSolution.java 151Reverse Words in a StringMediumSolution.java 162Find Peak ElementMediumSolution.java 167Two Sum II - Input Array Is SortedMediumSolution.java
No.ProblemLeetCode力扣PythonGoSolutionDifficultyTag 0017 Letter Combinations of a Phone Number LeetCode 力扣 Python CSDN Medium 回溯、暴力 0034 Find First and Last Position of Element in Sorted Array LeetCode 力扣 Python CSDN Medium 二分 0039 Combination Sum LeetCode 力扣 Python CSDN Medium 回溯 ...
class Solution { public int singleNumber(int[] nums) { int x = 0; for (int num : nums) // 1. 遍历 nums 执行异或运算 x ^= num; return x; // 2. 返回出现一次的数字 x } } 说明: 通过遍历数组中的每个数字,并使用异或运算将结果保存在result变量中,最终返回result即可。 C语言版本 #in...
// Solution 1: An interesting problem. As the elements are all integers, if not encountering 0, the absolute value of the product will continue to increase. Thus the product is either min or max. By keeping watch over both ends, we have the maximum product of subarrays. This holds only...