[LeetCode] 387. First Unique Character in a String

轉自LeetCode

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.

<Solution>

這題是要找沒有重複的字元中，在 string 中第一個出現的

第一種想法

• 準備一個 hash map，記錄所有字元出現的次數

• 按順序去歷遍 string，找到第一個出現次數為1的字元所在的 index

code 如下

C++

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

view raw firstUniqueCharInString_1.cpp hosted with ❤ by GitHub

第二種想法(參考資料)，會更快一點

• 用兩個指針，指針 cur 指在目前出現次數為1的字元，另一個指針 next 往前找

• 如果 next 指針有更新到 cur 指針指的字元，那麼將 cur 指到下一個出現次數為1的字元

• 如果 cur 指針已經指到尾了，那麼就回傳 -1

code 如下

C++

	class Solution {
	public:
	int firstUniqChar(string s) {
	if(s.empty()) {
	return -1;
	}
	int hashmap[256] = {0};
	int cur = 0, next = 1;
	const int len = s.length();
	hashmap[s[cur]]++;
	while(next < len) {
	hashmap[s[next]]++;
	//>> if current char shows more than one time,
	//>> then find next char which shows at first time
	while(cur < len && hashmap[s[cur]] > 1) {
	++cur;
	}
	if(cur == len) {
	return -1;
	}
	else if (hashmap[s[cur]] == 0) {
	//>> find a char which shows at first time
	hashmap[s[cur]]++;
	next = cur; //>> reset next pointer
	}
	++next;
	}
	return cur;
	}
	};

view raw firstUniqueCharInString_2.cpp hosted with ❤ by GitHub