LeetCode 389.找不同

389.找不同

思路一(较快)

将所给的字符串转换为字符数组,求字符数组的int和,作差,再转回char,返回

class Solution {
    public char findTheDifference(String s, String t) {
        char[] ss = s.toCharArray();
        char[] tt = t.toCharArray();
        int sums = 0;
        int sumt = 0;
        for(int i=0; i<ss.length; i++){
            sums += (int)ss[i];
        }
        for(int j=0; j<tt.length; j++){
            sumt += (int)tt[j];
        }
        return (char)(sumt-sums);
        
    }
}

思路二(慢)

使用sort对数据进行排序,然后依次对比

class Solution {
        public char findTheDifference(String s, String t) {
        char tempOne[] = s.toCharArray();
        char tempTwo[] = t.toCharArray();
        
        Arrays.sort(tempOne);
        Arrays.sort(tempTwo);

        for (int i = 0; i < tempOne.length; i ++) {
            if ((tempOne[i] - tempTwo[i]) != 0) {
                return tempTwo[i];
            }
        }

        return tempTwo[tempTwo.length - 1];
    }
}

思路三(最快)

异或逐个比较最后肯定剩下一个

class Solution {
    public char findTheDifference(String s, String t) {
        char[] s1=s.toCharArray();
        char[] t1=t.toCharArray();
        char result=0;
        for(int i=0;i<s1.length;i++)
        {
            result^=s1[i];
            result^=t1[i];
        }    
        result^=t1[t1.length-1];
        return result;
    }
}

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转载自blog.csdn.net/u014239185/article/details/86517553
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