# \#389 Find the Difference
## *給定字串s, t, t由s調換字元順序後組成, 並額外增加了一個字元, 找到並回傳這個多出來的字元*
## Log
- build 20210720 by syhuang

## XOR兩兩排除法
- 用XOR兩兩相消, 回傳消除到最後多出來的字元
- 搭配charCodeAt()和fromCharCode()對字元做運算

 var findTheDifference = function(s, t) {
    let result = 0;
    for(let i=0; i<s.length; i++){
        result ^= s[i].charCodeAt();
    }
    for(let i=0; i<t.length; i++){
        result ^= t[i].charCodeAt();
    }
    return String.fromCharCode(result);
};

## 解法2
- 用map下去做, 計算s字元出現次數後再用t去扣, 回傳沒找到or扣到0的字元
- leetcode submit runtime >95%, memory >50%

var findTheDifference = function(s, t) {
    let map = new Map();
    let result;
    s.split('').forEach(function(c){
        map.set(c,map.has(c)?map.get(c)+1:1);
    });
    try{
        t.split('').forEach(function(c){
            let count = map.get(c);
            if(!count){
                result = c;
                throw {};
            }else{
                map.set(c,count-1);
            }
        });
    }catch{};
    return result;
};

## 初見
- 用t下去逐字元跑迴圈, 對每個字元去除掉s中的相同字元, 當找不到字元時就表示是多餘字元
- leetcode submit結果memory過高(>100%), runtime 80%

var findTheDifference = function(s, t) {
    let oriStr = s;
    let result;
    try{
        t.split('').forEach(function(c){
            let cpos = oriStr.indexOf(c);
            if(cpos<0){
                result = c;
                throw {};
            }else oriStr = String.prototype.concat(oriStr.substr(0,cpos),oriStr.substr(cpos+1));
        });
    }catch{};
    return result;
};

## 備註
## 參考
- [三種參考解](https://leetcode.com/problems/find-the-difference/discuss/86904/3-Different-Python-Solutions-(Dictionary-Difference-XOR))
###### tags: `leetcode`, `leetcode-easy`