时间复杂度&&空间复杂度:O(n)(只扫一遍)&&O(1) class Solution { public: void sortColors(vector<int>& nums) { int l = 0,mid = 0,r = nums.size()-1; while(mid <= r){ if(nums[mid] == 0){ swap(nums[mid++],nums[l++]); } else if(nums[mid] == 2){ swap(nums[mid],num...
Leetcode 75 Sort Colors Given an array withnobjects colored red, white or blue, sort themin-placeso that objects of the same color are adjacent, with the colors in the order red, white and blue. Here, we will use the integers 0, 1, and 2 to represent the color red, white, and b...
Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue. Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively. void sort...
My Solutions to Leetcode problems. All solutions support C++ language, some support Java and Python. Multiple solutions will be given by most problems. Enjoy:) 我的Leetcode解答。所有的问题都支持C++语言,一部分问题支持Java语言。近乎所有问题都会提供多个算
i=j=0# 遍历两个数组,每次循环将较小的元素添加到已排序数组中whilei<len(left)andj<len(right):ifleft[i]<right[j]:sorted_array.append(left[i])i+=1else:sorted_array.append(right[j])j+=1# 收集左半部分剩余的元素,如果有的话whilei<len(left):sorted_array.append(left[i])i+=1# 收集右...
这道题也可以用一次三路快排。数组分为 3 部分,第一个部分都是 0,中间部分都是 1,最后部分都是 2 。 参考代码 packageleetcodefuncsortColors(nums[]int){zero,one:=0,0fori,n:=rangenums{nums[i]=2ifn<=1{nums[one]=1one++}ifn==0{nums[zero]=0zero++}}} ...
util.Arrays; public class ShellSort { public static void main(String []args) { int []arr ={1,4,2,7,9,8,3,6}; sort(arr); System.out.println(Arrays.toString(arr)); } public static void sort(int []arr) { // 增量gap,并逐步缩小增量 for(int gap = arr.length/2; gap > 0; ...
class Solution {public: void sortColors(int A[], int n) { int red = -1, white = -1, blue = -1; for (int i=0; i<n; else="" if="" pre=""></p>算法参考自:</p><p> https://leetcode.com/discuss/1827/anyone-with-one-pass-and-constant-space-solution</p><p> 算法二:红...
package leetcode import ( "sort" ) func frequencySort(s string) string { if s == "" { return "" } sMap := map[byte]int{} cMap := map[int][]byte{} sb := []byte(s) for _, b := range sb { sMap[b]++ } for key, value := range sMap { cMap[value] = append(cMap[va...
Explanation: First, we sort the values present at odd indices (1 and 3) in non-increasing order. So, nums changes from [4,1,2,3] to [4,3,2,1]. Next, we sort the values present at even indices (0 and 2) in non-decreasing order. So, nums changes from [4,1,2,3] to [2...