Likou 316. Remove duplicate letters C++

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Table of contents

Question requirements

Ideas

Likou submits AC code

C++ code

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Question requirements

• Remove some duplicate elements
• Keep the original order
• Minimize lexicographic order

Ideas

• Traverse a string and maintain a stack
• If the current character is already on the stack, skip it directly (see the program for details)
• When the top element of the stack > the current element , pop the top of the stack and then push it into the stack
  • If the top element of the stack is the last survivor, then refuse to pop out of the stack (the question requires that only duplicate characters can be deleted)
• When the top element of the stack < the current element , then it is pushed onto the stack smoothly.

If you still don't understand...

(Then it’s hopeless)

Then take a look at the program. There are comments in the program.

Likou submits AC code

class Solution {
public:
    string removeDuplicateLetters(string s) {
        vector<int> times(260, 0); //初始化 每一个字符在后续出现次数
        string sta = ""; //用string当作栈本体
        for (auto c : s) times[c] ++; //统计次数
        for (auto c : s) {
            //如果栈中已经有了，不给进，该字符次数--
            if (sta.find(c) != string::npos) {
                times[c] --;
                continue;
            }
            //栈非空           如果栈顶 > 当前元素 但是得保证必须是最后一个幸存者,才能出栈
            while(sta.size() && sta.back() > c && times[sta.back()])
                sta.pop_back();

            //新元素入栈
            sta.push_back(c);
            times[c] --; //次数--
        }
        return sta;
    }
};

In addition, it is also thoughtfully posted

C++ code

#include <iostream>
#include <string>
#include <vector>

using namespace std;

int main() {
    string s;
    cin >> s;
    vector<int> times(260, 0); //初始化 每一个字符在后续出现次数
    string sta = ""; //用string当作栈本体
    for (auto c : s) times[c] ++; //统计次数
    for (auto c : s) {
        //如果栈中已经有了，不给进，该字符次数--
        if (sta.find(c) != string::npos) {
            times[c] --;
            continue;
        }
        //栈非空 如果栈顶 > 当前元素 但是得保证必须是最后一个幸存者,才能出栈
        while(sta.size() && sta.back() > c && times[sta.back()])
            sta.pop_back();

        //新元素入栈
        sta.push_back(c);
        times[c] --; //次数--
    }
    cout << sta;
}

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