【LeetCode】154. Find Minimum in Rotated Sorted Array II (cpp),程序员大本营,技术文章内容聚合第一站。
189 Rotate Array 查看原文 旋转数组 给定一个数组,将数组中的元素向右移动k个位置,其中k是非负数。 示例1: 输入: [1,2,3,4,5,6,7] 和k=3输出: [5,6,7,1,2,3,4] 解释: 向右旋转1步: [7,1,2,3,4,5,6] 向右旋转2步: [6,7,1,2,3,4,5] 向右旋转3步: [5,6,7,1,2,3,4] ...
voidrotate(vector<int>&nums,int k){int s1=nums.size();k=k%s1;//如果nums=[1,2,3,4,5,6],k=11,我们要求余vector<int>temp(k,0);for(int i=0;i<k;i++)//把nums的后k位放在temp中temp[i]=nums[s1-k+i];for(int i=s1-k-1;i>=0;i--)//把nums的其余位往后挪k个位置nums[i+...
// 42_GetRotationArrayMinValue.cpp : Defines the entry point for the console application. // #include"stdafx.h" #include<iostream> usingnamespacestd; // Get min value from data[left] to data[right] intMinInOrder(int*data,intleft,intright) { intminValue = data[left]; for(inti = le...
题目描述: Given an array, rotate the array to the right byksteps, wherekis non-negative. Example 1: [1,2,3,4,5,6,7] [5,6,7,1,2,3,4] [7,1,2,3,4,5,6] [6,7,1,2,3,4,5] [5,6,7,1,2,3,4] Example 2:
第一次二分:找到最小数的位置,参考 find minimum number in rotated sorted array 第二次二分:确定 target 在左侧区间还是右侧,用一个普通的二分法即可找到。 class Solution: """ @param A: an integer rotated sorted array @param target: an integer to be searched @return: an integer """ def sear...
int minNumberInRotateArray(int* nums, int numsLen ) { int min=nums[0]; for (int i=0;i<numsLen-1;i++) { if ((nums[i]<=nums[i+1])&&(nums[i]<min)) min=nums[i]; } if (min<nums[numsLen-1]) return min; else return nums[numsLen-1]; } ...
Suppose a sorted array is rotated at some pivot unknown to you beforehand. 2.1 数组 5 (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search. If found in the array return its index, otherwise return -1. ...
0025-Reverse-Nodes-in-k-Group 0026-Remove-Duplicates-from-Sorted-Array 0027-Remove-Element 0028-Implement-strStr 0033-Search-in-Rotated-Sorted-Array/cpp-0033 CMakeLists.txt main.cpp 0034-Search-for-a-Range 0036-Valid-Sudoku 0037-Sudoku-Solver 0038-Count-and-Say 0039...
class Solution: """ @param A: an array @return: the maximum value of F(0), F(1), ..., F(n-1) """ def maxRotateFunction(self, A): # Write your code here s = sum(A) curr = sum(i*a for i, a in enumerate(A)) maxVal = curr for i in range(1, len(A)): curr +...