leetcode1347

class Solution:
    def minSteps(self, s: str, t: str) -> int:
        n = len(s)
        dic1 = {}
        for i in range(n):
            cur = s[i]
            if cur not in dic1:
                dic1[cur] = 1
            else:
                dic1[cur] += 1
        count = n
        for j in range(n):
            cur = t[j]
            if cur in dic1 and dic1[cur] > 0:
                dic1[cur] -= 1
                count -= 1
        return count

算法思路：hash，统计不同的字符个数。