zoukankan      html  css  js  c++  java
  • [leetcode-387-First Unique Character in a String]

    Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.

    Examples:

    s = "leetcode"
    return 0.
    
    s = "loveleetcode",
    return 2.
    

    Note: You may assume the string contain only lowercase letters.

    Brute force solution, traverse string s 2 times. First time, store counts of every character into the hash table,
    second time, find the first character that appears only once.
    if the string is extremely long, we wouldn't want to traverse it twice,
    so instead only storing just counts of a char, we also store the index, and then traverse the hash table.
     
    class Solution {
    public:
        int firstUniqChar(string s) {
            unordered_map<char, int> m;
            for(auto &c : s) {
                m[c]++;
            }
            for(int i = 0; i < s.size(); i++) {
                if(m[s[i]] == 1) return i;
            }
            return -1;
        }
    };

    class Solution {
    public:
        int firstUniqChar(string s) {
            unordered_map<char, pair<int, int>> m;
            int idx = s.size();
            for(int i = 0; i < s.size(); i++) {
                m[s[i]].first++;
                m[s[i]].second = i;
            }
            for(auto &p : m) {
                if(p.second.first == 1) idx = min(idx, p.second.second);
            }
            return idx == s.size() ? -1 : idx;
        }
    };

    参考:
    https://discuss.leetcode.com/topic/55082/c-2-solutions

  • 相关阅读:
    CCNA 6.9
    CCNA 6.5
    Google search
    CCNA 4.14 TP Correction
    CCNA 6.3
    CCNA 6.6
    有关 英语学习的一些网站
    法语学习笔记
    垃圾邮件分类(Scala 版本)
    SQL 面试经典问题 行列互相转化
  • 原文地址:https://www.cnblogs.com/hellowooorld/p/6965282.html
Copyright © 2011-2022 走看看