17. 电话号码的字母组合
2025-04-17 14:35:36

思路

这题我一开始使用的是BFS的思想,如下面的第一个代码所示。然而这样做存在不足,因为需要频繁构建StringBuilder来存储当前层的字符串。

对于字符串的频繁操作,如果使用DFS+回溯的思想,就可以减少这个反复创建的开销,效率更高。因为这个时候我们已经预先定好了将要插入字符的长度,并且可以重复利用之前的子字符串,只需要在到达叶子节点时把答案加入答案数组即可。

Code

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class Solution {
Queue<StringBuilder> ans = new LinkedList<>();
private static final char[][] MAP = {
{}, {}, {'a','b','c'}, {'d','e','f'}, {'g','h','i'},
{'j','k','l'}, {'m','n','o'}, {'p','q','r','s'},
{'t','u','v'}, {'w','x','y','z'}
};
public List<String> letterCombinations(String digits) {
List<String> ret = new ArrayList<>();
if (digits.isEmpty()) return ret;
ans.offer(new StringBuilder());
for (int i = 0; i < digits.length(); i++) {
bfs(digits.charAt(i) - '0');
}
while (!ans.isEmpty()) {
ret.add(ans.poll().toString());
}
return ret;
}

public void bfs(int num) {
int m = ans.size();
for (int i = 0; i < m; i++) {
StringBuilder sb = ans.poll();
for (int j = 0; j < MAP[num].length; j++) {
StringBuilder temp = new StringBuilder(sb);
ans.offer(temp.append(MAP[num][j]));
}
}
}
}
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class Solution {
private static final String[] MAPPING = new String[]{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};

private final List<String> ans = new ArrayList<>();
private char[] digits;
private char[] path;

public List<String> letterCombinations(String digits) {
int n = digits.length();
if (n == 0) {
return List.of();
}
this.digits = digits.toCharArray();
path = new char[n]; // 注意 path 长度一开始就是 n,不是空数组
dfs(0);
return ans;
}

private void dfs(int i) {
if (i == digits.length) {
ans.add(new String(path));
return;
}
for (char c : MAPPING[digits[i] - '0'].toCharArray()) {
path[i] = c; // 直接覆盖
dfs(i + 1);
}
}
}
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