}publicvoidreverse(int[] nums,intleft,intright){if(nums ==null|| nums.length == 1){return; }while(left <right){inttemp =nums[left]; nums[left]=nums[right]; nums[right]=temp; left++; right--; } } }
1.问题描述 给定一个 n × n 的二维矩阵 matrix 表示一个图像。请你将图像顺时针旋转 90 度。 你必须在 原地 旋转图像,这意味着你需要直接修改输入的二维矩阵。请不要 使用另一个矩阵来旋转图像。 2.测试用例 用例1: 输入:matrix = [[1,2,3],[4,5,6]
public int [] reverse(int[] arr, int left, int right){ if(arr == null || arr.length == 1) return arr; for(int i=0;i<(right-left+1)/2;i++){ int tmp = arr[left+i]; arr[left+i] = arr[right-i]; arr[right-i] = tmp; } return arr; } 1. 2. 3. 4. 5. 6. 7...
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 ...
Leetcode: Rotate Array 题目: Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4]. Note: Try to come up as many solutions as you can, there are at least 3 different...
classSolution{public:boolsearch(vector<int>&nums,int target){int size=nums.size();int left=0,right=size-1;while(left<=right){int mid=left+(right-left)/2;if(nums[mid]==target)returntrue;// 关键:对于重复元素nums[left] = nums[mid] = nums[right]难以判断target在哪个区间// 那就缩小[le...
void exchange(int* a, int* b){int n=*a;*a = *b;*b = n;} //交换a,b位置void reverse(int* nums,int left,int right){while(left<right){exchange(&nums[left],&nums[right]);left++;right--;} //对指定范围内元素进行翻转操作}void rotate(int* nums, int numsSize, int k){k%=nums...
给定一个整数数组nums,将数组中的元素向右轮转k个位置,其中k是非负数。 示例1: 输入:nums = [1,2,3,4,5,6,7], k = 3输出:[5,6,7,1,2,3,4]解释:向右轮转 1 步:[7,1,2,3,4,5,6]向右轮转 2 步:[6,7,1,2,3,4,5]向右轮转 3 步:[5,6,7,1,2,3,4] ...
/* * @lc app=leetcode id=33 lang=javascript * * [33] Search in Rotated Sorted Array *//** * @param {number[]} nums * @param {number} target * @return {number} */var search = function (nums, target) { // 时间复杂度：O(logn) // 空间复杂度：O(1) // [6,7,8,1,...
if nums[left] <= nums[middle] <= nums[right]: return nums[left] # 如果是连续递减,返回最右边的值 if nums[right] <= nums[middle] <= nums[left]: return nums[right] # 找到枢纽值的条件 if nums[middle] < nums[middle + 1] < nums[middle - 1]: ...