389. Find the Difference
2016-10-07 09:49
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原题链接: https://leetcode.com/problems/find-the-difference/
题目:
Given two strings s and t which consist of only lowercase letters.
String t is generated by random shuffling string s and then add one more letter at a random position.
Find the letter that was added in t.
给你两个小写字母组成的数组s和t,t是由s里面的字母乱序后再在某个随机位置加上一个另外的字母组成的,要你求出这个混进去的字母是什么。
如下:
解题思路:可以完全参照我之前的一道题的解法二,跳传送门过去看就好了 136. Single Number
AC代码:
网上找到的hash表做法
利用了unordered_map容器,是一个比java里hashmap的功能更广一点的容器。
即利用这个key-value存储的容器,按字符值来储存一个int值,如果该int值等于1,则只出现过一次。
题目:
Given two strings s and t which consist of only lowercase letters.
String t is generated by random shuffling string s and then add one more letter at a random position.
Find the letter that was added in t.
给你两个小写字母组成的数组s和t,t是由s里面的字母乱序后再在某个随机位置加上一个另外的字母组成的,要你求出这个混进去的字母是什么。
如下:
Input: s = "abcd" t = "abcde" Output: e Explanation: 'e' is the letter that was added.
解题思路:可以完全参照我之前的一道题的解法二,跳传送门过去看就好了 136. Single Number
AC代码:
class Solution { public: char findTheDifference(string s, string t) { char result=s[0]; for(int i=1;i<s.length();i++){ result^=s[i]; } for(int i=0;i<t.length();i++){ result^=t[i]; } return result; } }; //可以直接加起来,然后用for each循环遍历来进行异或运算,写简洁些 //class Solution { //public: // char findTheDifference(string s, string t) { // s += t; // int ch =0; // for(auto val: s) ch ^= val; // return ch; // } //};
网上找到的hash表做法
利用了unordered_map容器,是一个比java里hashmap的功能更广一点的容器。
即利用这个key-value存储的容器,按字符值来储存一个int值,如果该int值等于1,则只出现过一次。
class Solution { public: char findTheDifference(string s, string t) { unordered_map<char, int> hash; char ans; for(auto ch: s) hash[ch]++; for(auto ch: t) if(--hash[ch]<0) ans = ch; return ans; } };
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