LeetCode75-1657-Close-Two-String

🧠 思路总结

判断两个字符串是否 close：

1. 长度是否相同
2. 是否包含完全相同的字母集合
3. 字母出现次数是否只是一个排列

需要找什么可以一直不变

• Operation 1: Swap any two existing characters.
  For example, abcde -> aecdb。本质上是需要确保word1中有的charword2中也得有
• Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character. For example, aacabb -> bbcbaa (all a’s turn into b’s, and all b’s turn into a’s)。本质上是需要确保word1中有的charword2中也得有 并且 需要确保occurrence排序后一模一样

✅ 代码（Java）

class Solution {
    public boolean closeStrings(String word1, String word2) {
        if (word1.length() != word2.length()) {
            return false;
        }
        Map<Character, Integer> m1 = new HashMap<>();
        Map<Character, Integer> m2 = new HashMap<>();
        for (char c : word1.toCharArray()) {
            m1.put(c, m1.getOrDefault(c, 0) + 1);
        }
        for (char c : word2.toCharArray()) {
            m2.put(c, m2.getOrDefault(c, 0) + 1);
        }
        if (!m1.keySet().equals(m2.keySet())) {
            return false;
        }
        List<Integer> occ1 = new ArrayList<>(m1.values());
        List<Integer> occ2 = new ArrayList<>(m2.values());
        Collections.sort(occ1);
        Collections.sort(occ2);
        return occ1.equals(occ2);
    }
}