代码随想录 Day25 - 回溯(二)
作者:jjn0703
- 2023-07-23 江苏
本文字数:1635 字
阅读完需:约 5 分钟
216. 组合总和 III
关键点在于 startIndex,通过该变量控制回溯的决策阶段。
package jjn.carl.backtrack;
import java.util.ArrayList;
import java.util.List;
/**
* @author Jiang Jining
* @since 2023-07-23 22:19
*/
public class LeetCode216 {
public List<List<Integer>> combinationSum3(int k, int n) {
List<List<Integer>> result = new ArrayList<>();
backtrack(k, n, new ArrayList<>(), result, 0, 1);
return result;
}
private void backtrack(int k, int n, List<Integer> path, List<List<Integer>> result, int sum, int startIndex) {
if (path.size() > k || sum > n) {
return;
}
if (path.size() == k && sum == n) {
result.add(new ArrayList<>(path));
return;
}
for (int i = startIndex; i < 10; i++) {
path.add(i);
sum += i;
backtrack(k, n, path, result, sum, i + 1);
path.remove(path.size() - 1);
sum -= i;
}
}
public static void main(String[] args) {
System.out.println("new LeetCode216().combinationSum3(3, 7) = " + new LeetCode216().combinationSum3(3, 7));
System.out.println("new LeetCode216().combinationSum3(3, 9) = " + new LeetCode216().combinationSum3(3, 9));
}
}
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17. 电话号码的字母组合
构造一个 KV 键值对,把每个键及对应的字母关系存储起来,随后回溯,每一个阶段取输入对应的可选项。
package jjn.carl.backtrack;
import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
/**
* @author Jiang Jining
* @since 2023-07-23 22:43
*/
public class LeetCode17 {
private static final Map<Character, List<String>> KEY_MAP = Map.of('2', List.of("a", "b", "c"),
'3', List.of("d", "e", "f"), '4', List.of("g", "h", "i"),
'5', List.of("j", "k", "l"), '6', List.of("m", "n", "o"),
'7', List.of("p", "q", "r", "s"), '8', List.of("t", "u", "v"),
'9', List.of("w", "x", "y", "z"));
public List<String> letterCombinations(String digits) {
List<String> result = new ArrayList<>();
if (digits.length() == 0) {
return result;
}
backtrack(digits, new StringBuilder(), result, 0);
return result;
}
private void backtrack(String digits, StringBuilder stringBuilder, List<String> result, int startIndex) {
if (stringBuilder.length() > digits.length()) {
return;
}
if (stringBuilder.length() == digits.length()) {
result.add(stringBuilder.toString());
return;
}
for (int i = startIndex; i < digits.length(); i++) {
char c = digits.charAt(i);
List<String> options = KEY_MAP.get(c);
for (String option : options) {
stringBuilder.append(option);
backtrack(digits, stringBuilder, result, i + 1);
stringBuilder.deleteCharAt(stringBuilder.length() - 1);
}
}
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String digits = scanner.nextLine();
List<String> result = new LeetCode17().letterCombinations(digits);
System.out.println("result = " + result);
}
}
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版权声明: 本文为 InfoQ 作者【jjn0703】的原创文章。
原文链接:【http://xie.infoq.cn/article/3c50b1f81776aae485bc47170】。
本文遵守【CC-BY 4.0】协议,转载请保留原文出处及本版权声明。
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