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  • [LeetCode] 316. Remove Duplicate Letters 移除重复字母

    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"

    Credits:
    Special thanks to @dietpepsi for adding this problem and creating all test cases.

    给定一个只含有小写字母的字符串,移除重复的字符,每种字符只出现1次,返回字符串是按字典顺序中最小的组合。

    解法:贪心算法Greedy,用1个HashMap或者数组统计出现的字符。然后再遍历一遍数组, 假设position = 0,每次找到字符比position的小就更新position,同时也更新count,当count[i] == 0的时候,这个字符就是我们要找的结果字符串里的第一个字符。之后因为其他字符的count还都 > 1,继续在s.substring(position + 1)的子串里递归查找第二个字符,注意要在这个子串里把第一个字符清除掉。

    Java:

    public class Solution {
        public String removeDuplicateLetters(String s) {
            if(s == null || s.length() == 0) {
                return s;
            }
            
            int[] count = new int[26];
            char[] res = new char[26];
            int len = s.length();
            for(int i = 0; i < s.length(); i++) {
                count[s.charAt(i) - 'a']++; 
            }
            
            int pos = 0;
            for(int i = 0; i < len; i++) {
                if(s.charAt(i) < s.charAt(pos)) {
                    pos = i;
                }
                count[s.charAt(i) - 'a']--;     
                if(count[s.charAt(i) - 'a'] == 0) {         // found first minimum char
                    break;
                }
            }
            
            String charToRemove = String.valueOf(s.charAt(pos));
            return charToRemove + removeDuplicateLetters(s.substring(pos + 1).replaceAll(charToRemove, ""));
        }
    }  

    Python:

    class Solution(object):
        def removeDuplicateLetters(self, s):
            """
            :type s: str
            :rtype: str
            """
            remaining = collections.defaultdict(int)
            for c in s:
                remaining[c] += 1
    
            in_stack, stk = set(), []
            for c in s:
                if c not in in_stack:
                    while stk and stk[-1] > c and remaining[stk[-1]]:
                        in_stack.remove(stk.pop())
                    stk += c
                    in_stack.add(c)
                remaining[c] -= 1
            return "".join(stk)  

    C++:

    class Solution {
    public:
        string removeDuplicateLetters(string s) {
            int m[256] = {0}, visited[256] = {0};
            string res = "0";
            for (auto a : s) ++m[a];
            for (auto a : s) {
                --m[a];
                if (visited[a]) continue;
                while (a < res.back() && m[res.back()]) {
                    visited[res.back()] = 0;
                    res.pop_back();
                }
                res += a;
                visited[a] = 1;
            }
            return res.substr(1);
        }
    }; 

    All LeetCode Questions List 题目汇总

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  • 原文地址:https://www.cnblogs.com/lightwindy/p/8564419.html
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