其中remove函数需要使用algorithm库文件。算法复杂度为O(26 * n) = O(n) 1classSolution {2public:3stringremoveDuplicateLetters(strings) {4vector<int> alp(26, -1);5for(inti =0, n = s.size(); i < n; i++)6alp[(int)(s[i] -'a')] =i;7intpos = -1;8for(inti =0, n = s.s...
Can you solve this real interview question? Remove All Adjacent Duplicates in String II - You are given a string s and an integer k, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them, causing the left and th
[LeetCode] 1209. Remove All Adjacent Duplicates in String II 移除字符串中所有相邻的重复字符之二 You are given a stringsand an integerk, akduplicate removal consists of choosingkadjacent and equal letters fromsand removing them, causing the left and the right side of the deleted substring to c...
代码(Go) func removeDuplicateLetters(s string) string { // lastIndex[ch] 表示 ch 在 s 中的最后一个出现的位置 lastIndex := make(map[rune]int) // 带下标遍历 s 中的字符 for i, ch := range s { // 更新每个字符最后一次出现的位置 lastIndex[ch] = i } // isInStack[ch] 表示 ch ...
* https://discuss.leetcode.com/topic/31404/a-short-o-n-recursive-greedy-solution * 贪心算法 * */ public class Solution { public String removeDuplicateLetters(String s) { if(s==null || s.length()<=0) return ""; else { int[] count=new int[26]; ...
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leetcode刷题指南之RemoveDuplicateLetters 编辑| 刘凯旋 公众号 | 转载自 【编程与算法之美】 1.题目分析 给定一个字符串,删去重复字母使得每个字母只剩一个,并且字典序最小。 2.解题思路 最暴力的方法就是用dfs去搜索,记录哪些字母还没有出现,这个的复杂度是O(26!)的 ,显然不行。
3.代码示例 1classSolution 2{ 3public: 4stringremoveDuplicateLetters(strings) 5{ 6intn=s.size(); 7vector<int>c(26,0),in(26,0); 8for(inti=0;i<n;i++) ++c[s[i]-'a']; 9stringans; 10for(inti=0;i<n;i++) 11{ 12--c[s[i...
Given sorted array A =[1,1,1,2,2,3], Your function should return length =5, and A is now[1,1,2,2,3]. 典型的两指针问题 两个指针指向初始位置,一个指针i开始遍历,记录出现相同数的个数 如果遍历的指针i等于其前面的指针index且cnt个数超过两个,则继续移动遍历的指针 ...
Can you solve this real interview question? Remove Duplicate Letters - Given a string s, remove duplicate letters so that every letter appears once and only once. You must make sure your result is the smallest in lexicographical order among all possible