交换相邻两数 如果只是交换相邻两数,那么最少交换次数为该序列的逆序数. 交换任意两数 数字的总个数减去循环节的个数?? A cycle is a set of elements, each of which is in the place of another.  So in example sequences { 2, 1, 4, 3}, there are two cycles: {1, 2} and {3, 4}.  1 is in the place where 2 needs to G

Given a non-empty integer array, find the minimum number of moves required to make all array elements equal, where a move is incrementing a selected element by 1 or decrementing a selected element by 1. You may assume the array's length is at most 10

原题链接在这里:https://leetcode.com/problems/minimum-moves-to-equal-array-elements-ii/ 题目: Given a non-empty integer array, find the minimum number of moves required to make all array elements equal, where a move is incrementing a selected element by 1 or d

#462.   Minimum Moves to Equal Array Elements II Given a non-empty integer array, find the minimum number of moves required to make all array elements equal, where a move is incrementing a selected element by 1 or decrementing a selected element by 1

922. Sort Array By Parity II 题目描述 Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even. Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even. You may re

Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even. Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even. You may return any answer array that satisfi

［抄题］: Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1. Example: Input: [1,2,3] Output: 3 Explanation: Only three moves are needed

Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even. Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even. You may return any answer array that satisfi

problem 905. Sort Array By Parity solution1: class Solution { public: vector<int> sortArrayByParity(vector<int>& A) { vector<int> even, odd; for(auto a:A) { ==) even.push_back(a); else odd.push_back(a); } even.insert(even.end(), odd.

problem 922. Sort Array By Parity II solution1: class Solution { public: vector<int> sortArrayByParityII(vector<int>& A) { int n = A.size(); , j=; i<n, j<n; ) { ==) i += ; ==) j += ; if(j<n) swap(A[i], A[j]); } return A; } }; 参考 1

Follow up for "Find Minimum in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might be

Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 这道寻找旋转有序数组的最小值肯定不能通过直接遍历整个数组来寻找,这个方法过于简单粗暴,这样的话,旋不

题目概述: Suppose a sorted array is rotated at some pivot unknown to you beforehand.(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).Find the minimum element.You may assume no duplicate exists in the array. 解题思路: 在一个有顺序的数组中查找个最小值,很容易想到的就是二分法,的确,这里用的就是二分,

原题链接在这里:https://leetcode.com/problems/minimum-moves-to-equal-array-elements/ 题目: Given a non-empty integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1.

1. 画图, 直观. 2. 讨论数组为空或者个数为零. 3. 讨论首尾, 若为翻转过的则进行查找直到最后两个数进行比较, 取小者. public class Solution { /** * @param num: a rotated sorted array * @return: the minimum number in the array */ public int findMin(int[] num) { // write your code here if(num == null ||

Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. 二分查找就行了. class Solution { public: int findMin(vect

原题链接在这里:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/ 题目: Follow up for "Find Minimum in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted arra

原题链接在这里:https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/ Method 1 就是找到第一个违反升序的值,就是最小值,若是没有,那么第一个值就是最小值. Time O(n). Method 2 Binary Search. 与Find Peak Element类似. while 的条件是 l < r 并且 nums[l] > nums[r], 不然 l 到 r 这一段就是sorted的. nums[mi

Follow up for "Find Minimum in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might be

Follow up for "Find Minimum in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might be

多写限制条件可以加快调试速度. ======= Follow up for "Find Minimum in Rotated Sorted Array":What if duplicates are allowed? Would this affect the run-time complexity? How and why? Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e.