389. Find the Difference Add to List(新添字符)

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Total Accepted: 60224
Total Submissions: 117313
Difficulty: Easy
Contributor: LeetCode
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.

Example:

Input:
s = “abcd”
t = “abcde”

Output:
e

Explanation:
‘e’ is the letter that was added.

以前有个字符串,再给一个新的字符串,看新加的是哪一个

解法1:

class Solution {public:    char findTheDifference(string s, string t) {    bool hap[26] ={ true};    for (int i = 0; i < t.size(); i++)    {        hap[t[i] - 'A' - 32]=false;    }    for (int i = 0; i < s.size(); i++)    {        hap[s[i] - 'A'-32]=true;    }    for (int i = 0; i < 26; i++)    {        if (hap[i] == false)        {        return char(i + 'A' + 32) ;        }    }    }};//很稳健的超时了,下面是大神的

解法2:

class Solution {public:    char findTheDifference(string s, string t) {    unordered_map<char, int>a;//当然建立哈希表啊,老哥,char是索引类型,int为元素类型    for (char b : s)//遍历短的字符串    {        a[b]++;    }    for (char b : t)    {        if( --a[b]<0)//多余的那个自然就多减了        {            cout << b;//        }    }    }

解法3://运用按位异或解决:   a*a=0;并且符合交换律class Solution {public:    char findTheDifference(string s, string t) {    char res=0;    for(char a:s)    {        res^=a;    }    for(char b:t)    {        res^=b;    }    return res;    }

加减法.............class Solution {public:    char findTheDifference(string s, string t) {    char res = 0;        for (char c : s) res -= c;        for (char c : t) res += c;       return res;    }};

解法4:class Solution {public:    char findTheDifference(string s, string t) {        return accumulate(begin(s), end(s += t), 0, bit_xor<int>());    }};

参考链接:
http://www.cnblogs.com/grandyang/p/5816418.html