【leetcode】1247. Minimum Swaps to Make Strings Equal

题目如下：

解题思路：这个题目还是有点考巧劲的。如果s1[i] != s2[i]，只有这两种取值 s1[i] = x,s2[i] = y(记为xy)和s1[i] = y,s2[i] = x(记为yx)。理想情况下，进行s1[i]和s2[j]的交换，使得s1[i] = s2[i] 与 s1[j] = s2[j]同时成立，那么一定是交换次数最少的场景。而题目中例子1的交换则符合这个场景，只要分别求出(xy)和(yx)的个数，然后分别进行(xy)和(yx)的组内交换，剩余的配对再进行例子2的组间交换，即可求得最小的交换次数。

代码如下：

class Solution(object):
    def minimumSwap(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: int
        """
        dic = {}
        dic[('x','y')] = 0
        dic[('y','x')] = 0

        for (i,j) in zip(s1,s2):
            if i != j:
                dic[(i,j)]  += 1
        if (dic[('x','y')] + dic[('y','x')] ) % 2 != 0:
            return -1
        elif dic[('x','y')] % 2 == 0 and dic[('y','x')] % 2 == 0:
            return dic[('x','y')] / 2  + dic[('y','x')] / 2
        return dic[('x','y')] / 2  + dic[('y','x')] / 2 + 2