README.md

859. 亲密字符串

English Version

题目描述

给你两个字符串 s 和 goal ，只要我们可以通过交换 s 中的两个字母得到与 goal 相等的结果，就返回 true ；否则返回 false 。

交换字母的定义是：取两个下标 i 和 j （下标从 0 开始）且满足 i != j ，接着交换 s[i] 和 s[j] 处的字符。

• 例如，在 "abcd" 中交换下标 0 和下标 2 的元素可以生成 "cbad" 。

 

示例 1：

输入：s = "ab", goal = "ba"
输出：true
解释：你可以交换 s[0] = 'a' 和 s[1] = 'b' 生成 "ba"，此时 s 和 goal 相等。

示例 2：

输入：s = "ab", goal = "ab"
输出：false
解释：你只能交换 s[0] = 'a' 和 s[1] = 'b' 生成 "ba"，此时 s 和 goal 不相等。

示例 3：

输入：s = "aa", goal = "aa"
输出：true
解释：你可以交换 s[0] = 'a' 和 s[1] = 'a' 生成 "aa"，此时 s 和 goal 相等。

 

提示：

• 1 <= s.length, goal.length <= 2 * 104
• s 和 goal 由小写英文字母组成

解法

Python3

class Solution:
    def buddyStrings(self, s: str, goal: str) -> bool:
        m, n = len(s), len(goal)
        if m != n:
            return False
        diff = sum(1 for i in range(n) if s[i] != goal[i])
        cnt1, cnt2 = Counter(s), Counter(goal)
        if cnt1 != cnt2:
            return False
        return diff == 2 or (diff == 0 and any(e > 1 for e in cnt1.values()))

Java

class Solution {
    public boolean buddyStrings(String s, String goal) {
        int m = s.length(), n = goal.length();
        if (m != n) {
            return false;
        }
        int diff = 0;
        int[] cnt1 = new int[26];
        int[] cnt2 = new int[26];
        for (int i = 0; i < n; ++i) {
            int a = s.charAt(i), b = goal.charAt(i);
            ++cnt1[a - 'a'];
            ++cnt2[b - 'a'];
            if (a != b) {
                ++diff;
            }
        }
        boolean f = false;
        for (int i = 0; i < 26; ++i) {
            if (cnt1[i] != cnt2[i]) {
                return false;
            }
            if (cnt1[i] > 1) {
                f = true;
            }
        }
        return diff == 2 || (diff == 0 && f);
    }
}

C++

class Solution {
public:
    bool buddyStrings(string s, string goal) {
        int m = s.size(), n = goal.size();
        if (m != n) return false;
        int diff = 0;
        vector<int> cnt1(26);
        vector<int> cnt2(26);
        for (int i = 0; i < n; ++i)
        {
            ++cnt1[s[i] - 'a'];
            ++cnt2[goal[i] - 'a'];
            if (s[i] != goal[i]) ++diff;
        }
        bool f = false;
        for (int i = 0; i < 26; ++i)
        {
            if (cnt1[i] != cnt2[i]) return false;
            if (cnt1[i] > 1) f = true;
        }
        return diff == 2 || (diff == 0 && f);
    }
};

Go

func buddyStrings(s string, goal string) bool {
	m, n := len(s), len(goal)
	if m != n {
		return false
	}
	diff := 0
	cnt1 := make([]int, 26)
	cnt2 := make([]int, 26)
	for i := 0; i < n; i++ {
		cnt1[s[i]-'a']++
		cnt2[goal[i]-'a']++
		if s[i] != goal[i] {
			diff++
		}
	}
	f := false
	for i := 0; i < 26; i++ {
		if cnt1[i] != cnt2[i] {
			return false
		}
		if cnt1[i] > 1 {
			f = true
		}
	}
	return diff == 2 || (diff == 0 && f)
}

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