zoukankan      html  css  js  c++  java
  • leetcode 316. Remove Duplicate Letters

    Given a string which contains only lowercase letters, remove duplicate letters so that every letter appear once and only once. You must make sure your result is the smallest in lexicographical order among all possible results.

    Example:

    Given "bcabc"
    Return "abc"

    Given "cbacdcbc"
    Return "acdb"

    题目意思:去掉重复的字符,保持元序列的位置,并使得字典序最小

    真的是。。。想了半天也没想到用栈。看了他人的代码,还有用string 直接搞的

    class Solution {
    public:
        string removeDuplicateLetters(string s) {
            int n = s.length();
            vector<int> cnt(26);
            vector<int> vis(26);
            for (int i = 0; i < n; ++i) {
                int x = s[i] - 'a';
                cnt[x]++;
            }
            stack<int> st;
            for (int i = 0; i < n; ++i) {
                int x = s[i] - 'a';
                cnt[x]--;
                if (st.empty()) {
                    st.push(x);
                    vis[x] = 1;
                }
                else {
                    while (!st.empty() && st.top() > x && cnt[st.top()] && !vis[x]) {
                        vis[st.top()] = 0;
                        st.pop();
                    }
                    if (!vis[x]) st.push(x),vis[x] = 1;
                }
            }
            string t = "";
            while (!st.empty()) {
                int y = st.top();
                st.pop();
                t = char(y + 'a') + t;
            }
            return t;
        }
    };
    class Solution {
    public:
        string removeDuplicateLetters(string s) {
            vector<int> dict(256, 0);
            vector<bool> visited(256, false);
            for(auto ch : s)  dict[ch]++;
            string result = "0";
            /** the key idea is to keep a monotically increasing sequence **/
            for(auto c : s) {
                dict[c]--;
                /** to filter the previously visited elements **/
                if(visited[c])  continue;
                while(c < result.back() && dict[result.back()]) {
                    visited[result.back()] = false;
                    result.pop_back();
                }
                result += c;
                visited[c] = true;
            }
            return result.substr(1);
        }
    };
  • 相关阅读:
    achivemq(消息队列)的使用
    java高并发当时处理的思路
    字符串的应用
    正则表达式
    文本文件的读取与写入
    继承
    冒泡排序法
    类与对象
    数据类型
    关键字和语句
  • 原文地址:https://www.cnblogs.com/pk28/p/7300929.html
Copyright © 2011-2022 走看看