69 lines
2.4 KiB
Markdown
69 lines
2.4 KiB
Markdown
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---
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id: "202203102"
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date: "2022/03/10 21:15"
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title: "Q40 组合总和2(combination-sumⅡ)"
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tags: ["java", "leetcode", "dfs"]
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index_img: https://qiniupic.fleyx.com/blog/202203101631050.jpg?imageView2/2/w/200
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banner_img: https://qiniupic.fleyx.com/blog/202203101631050.jpg
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categories:
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- "算法"
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- "leetcode刷题"
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---
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<span id="blogIdSpan" style="display:none">202203102</span>
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## 解析思路
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leetcode 中等难度，题目描述[点击这里](https://leetcode-cn.com/problems/combination-sum-ii/)。
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本题属于[组合排序](https://blog.fleyx.com/blog/detail/202203101/)的进阶题型。建议先看[上一篇](https://blog.fleyx.com/blog/detail/202203101/)的解析。
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本题跟上一题区别只有一点是**选择的数字不能重复**,如何实现不重复呢？
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很简单，只需要将传入的 index 设置成 index+1，同时 for 循环时每次都循环到下一个不同的元素上
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这样就不会选取到已经选过的元素。
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代码如下：
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## 代码
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```java
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public class Q40 {
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public List<List<Integer>> combinationSum2(int[] candidates, int target) {
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List<List<Integer>> res = new ArrayList<>();
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Arrays.sort(candidates);
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dfs(candidates, target, 0, new Stack<>(), res);
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return res;
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}
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private void dfs(int[] candidates, int target, int index, Stack<Integer> temp, List<List<Integer>> res) {
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if (target == 0) {
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//说明找到一个结果序列
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res.add(new ArrayList<>(temp));
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return;
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}
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for (int i = index; i < candidates.length; ) {
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if (candidates[i] > target) {
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//前面已经排序过，所以在这里可以进行剪枝操作，如果candidates[index]都小于target了，那就不需要比较后面的了，肯定不满足要求
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return;
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}
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temp.push(candidates[i]);
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dfs(candidates, target - candidates[i], i + 1, temp, res);
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temp.pop();
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//手动控制i的增长，对于同一个数字不能重复处理
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int nextI = i + 1;
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while (nextI < candidates.length && candidates[nextI] == candidates[i]) {
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nextI++;
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}
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i = nextI;
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}
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}
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public static void main(String[] args) {
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new Q40().combinationSum2(new int[]{10, 1, 2, 7, 6, 1, 5}, 8).forEach(System.out::println);
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}
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}
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```
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