【LeetCode】387. First Unique Character in a String 解題報告

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387. First Unique Character in a String / Easy

Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.

Example 1:

Input: s = “leetcode”
Output: 0

Example 2:

Input: s = “loveleetcode”
Output: 2

Example 3:

Input: s = “aabb”
Output: -1

Comstraints

  • 1 <= s.length <= 10^5
  • s consists of only lowercase English letters.

Solution 1: Brute Force

思路

暴力解，利用雙迴圈不斷比對。用在這題不會 TLE。

效能

Complexity

  • Time Complexity: O(N^2)
  • Space Complexity: O(1)

LeetCode Result

Code

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class Solution {
public:
    int firstUniqChar(string s) {
        int cnt = 0;
        for(int i = 0; i < s.size(); ++i) {
            char now_char = s[i];
            bool found = false;
            for(int j = 0; j < s.size(); ++j) {
                if(i == j) continue;
                if(now_char == s[j]) {
                    found = true;
                    break;
                } 
            }
            if(!found) return i;
        }
        return -1;
    }
};

Solution 2: Hash Map

思路

第一次迴圈遍歷整個字串，利用 Hash Map 把每個字元的出現次數都記錄下來。
第二次迴圈再次遍歷字串，如果 Hash Map 中紀錄的 value 為 1，便回傳該字元的 index。
如果所有字元都出現兩次以上就回傳 -1。

但這題貌似用 Hash Map 卻變慢了…

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    int firstUniqChar(string s) {
        unordered_map<char, int> umap;
        for(int i = 0; i < s.size(); ++i) {
            ++umap[s[i]];
        }
        
        for(int i = 0; i < s.size(); ++i) {
            if(umap[s[i]] == 1) return i;
        }
        return -1;
    }
};

Solution 3: Hash Map (Array)

思路

和上一個解法一樣，只是換成用 Array 代替 unordered_map 實現簡單的 Hash Map，速度上快了不少。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    int firstUniqChar(string s) {
        short hmap[26] = {0};
        const int n = s.size();
        for(const auto& c: s) ++hmap[c-'a'];
        for(int i = 0; i < n; ++i) {
            if(hmap[s[i]-'a'] == 1) return i;
        }
        return -1;
    }
};