给定两个字符串 s1
和 s2
,请编写一个程序,确定其中一个字符串的字符重新排列后,能否变成另一个字符串。
示例 1:
输入:s1
= "abc",s2
= "bca" 输出: true
示例 2:
输入:s1
= "abc",s2
= "bad" 输出: false
说明:
0 <= len(s1) <= 100
0 <= len(s2) <= 100
用一个哈希表作为字符计数器,O(n)
时间内解决。
class Solution:
def CheckPermutation(self, s1: str, s2: str) -> bool:
n1, n2 = len(s1), len(s2)
if n1 != n2:
return False
counter = collections.Counter()
for i in range(n1):
counter[s1[i]] += 1
counter[s2[i]] -= 1
for val in counter.values():
if val != 0:
return False
return True
class Solution {
public boolean CheckPermutation(String s1, String s2) {
int n1 = s1.length(), n2 = s2.length();
if (n1 != n2) {
return false;
}
Map<Character, Integer> counter = new HashMap<>();
for (int i = 0; i < n1; ++i) {
char c1 = s1.charAt(i), c2 = s2.charAt(i);
counter.put(c1, counter.getOrDefault(c1, 0) + 1);
counter.put(c2, counter.getOrDefault(c2, 0) - 1);
}
for (int val : counter.values()) {
if (val != 0) {
return false;
}
}
return true;
}
}