voidrotate(vector<int>& nums,intk) {intn =nums.size();if(n ==0|| k ==0)return;//求最大公约数inta = n, c = k, b = a %c;while(b !=0) { a=c; c=b; b= a %c; }//依次替换while(c--) {intoldNum =nums[c];intcurPos = (c + k) %n;while(curPos !=c) {intcu...
Array.ConstrainedCopy(zong, 0, nums,0 ,nums.Length );修改为一个for循环: for(inti=0;i<nums.Length;i++) { nums[i]=zong[i]; } 这样也可以Accepted! 这里还有一个我至今没有看懂的方法,先放在这里,期待以后可以看懂: publicclassSolution {publicvoidRotate(int[] nums,intk) {intsz,n,temp; ...
题目链接: Rotate Array : leetcode.com/problems/r 轮转数组: leetcode-cn.com/problem LeetCode 日更第 86 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满 发布于 2022-04-14 09:29 力扣(LeetCode) Python 算法 赞同添加评论 分享喜欢收藏申请转载 ...
Can you solve this real interview question? Rotate Array - Given an integer array nums, rotate the array to the right by k steps, where k is non-negative. Example 1: Input: nums = [1,2,3,4,5,6,7], k = 3 Output: [5,6,7,1,2,3,4] Explanation: rotate 1
LeetCode Rotate Array 翻转数组 题意:给定一个数组,将该数组的后k位移动到前n-k位之前。(本题在编程珠玑中第二章有讲) 思路: 方法一:将后K位用vector容器装起来,再移动前n-k位到后面,再将容器内k位插到前面。 1 class Solution { 2 public:...
1、先得到有效的k, 2、反转所有数组 3、反转下标从0到k的数组 4、 反转下标从k到length - 1的数组 3、代码实现 class Solution { public void rotate(int[] nums, int k) { if (nums == null || nums.length == 0 || k <= 0) return; ...
Can you solve this real interview question? Search in Rotated Sorted Array - There is an integer array nums sorted in ascending order (with distinct values). Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1
Solution: 三步翻转法 classSolution{publicvoidrotate(int[]nums,intk){//三次翻转法if(nums==null||nums.length==0){return;}if(k>nums.length){k=k%nums.length;}intdiff=nums.length-k;if(diff<0)return;// 3 steps reversereverse(nums,0,diff-1);// reverse first part: 0 ~ (length - k...
nums is an ascending array that is possibly rotated. -104 <= target <= 104 思考 从题目提及的时间复杂度,可以知道这次肯定是需要用到二分法的。 我们先分析一下旋转之后(r1,r2)->(r2,r1)的列表。 [4,5,6,7,0,1,2] 这样一个列表,我们可以知道,因为是将升序列表的r2部分移植到了r1部分,所以移植...
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