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 ways to solve this pro…

Given an array, rotate the array to the right by k steps, where k is non-negative. Example 1: Input: [1,2,3,4,5,6,7] and k = 3 Output: [5,6,7,1,2,3,4] Explanation: rotate 1 steps to the right: [7,1,2,3,4,5,6] rotate 2 steps to the right: [6,7,1,2,3,4…

Given an array, rotate the array to the right by k steps, where k is non-negative. 给定一个数组,并且给定一个非负数的值k, 把数组往右旋转k步,要求不返回新的数组,直接改变原数组 例子1: 给定数组: [1,2,3,4,5,6,7] 给定 k = 3 输出数组: [5,6,7,1,2,3,4] 解析: 往右旋转1步: [7,1,2,3,4,5,6] 往右旋转2步: [6,7,1,2,3,4,5] 往右旋转3步:…

class Solution {public: void rotate(vector<int>& nums, int k) { int n=nums.size(); int i=0; //------------------- //解法一  会超时 //-------------------- k=k%n; while(i<k){ int temp=nums[n-1]; for(int j=n-1;j>0;j--){ nums[j]=nums[j-1]; } nums[0]…

将包含 n 个元素的数组向右旋转 k 步.例如,如果  n = 7 ,  k = 3,给定数组  [1,2,3,4,5,6,7]  ,向右旋转后的结果为 [5,6,7,1,2,3,4].注意:尽可能找到更多的解决方案,这里最少有三种不同的方法解决这个问题. 详见:https://leetcode.com/problems/rotate-array/description/ Java实现: 方法一: class Solution { public void rotate(int[] nums, i…

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. class Solution { public: void rotate(int nums[], int n, int k) { || k==n ) return; k %= n; vector<int> cha; cha.reserve(k); int i; for(i=n-k; i<n; i++)…

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 ways to solve this pr…

You are given an n x n 2D matrix representing an image. Rotate the image by 90 degrees (clockwise). Note:You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the…

一句话思路:反正只是寻找一个最小区间,断开也能二分.根据m第一次的落点,来分情况讨论. 一刷报错: 结构上有根本性错误:应该是while里面包括if,不然会把代码重复写两遍,不好. //situation1 if (nums[mid] > nums[start]) { while (start + 1 < mid) { mid = start + (end - start) / 2; if (nums[mid] == target) { return mid; } else if (nums[…

假设一个按照升序排列的有序数组从某未知的位置旋转.(比如 0 1 2 4 5 6 7 可能变成 4 5 6 7 0 1 2).找到其中最小的元素.你可以假设数组中不存在重复的元素.详见:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/description/ Java实现: class Solution { public int findMin(int[] nums) { int low=0; int high…