四刷40

Author: diguage

docs/0040-combination-sum-ii.adoc

@@ -1,40 +1,45 @@
 [#0040-combination-sum-ii]
-= 40. Combination Sum II
+= 40. 组合总和 II
 
-{leetcode}/problems/combination-sum-ii/[LeetCode - Combination Sum II^]
+https://leetcode.cn/problems/combination-sum-ii/[LeetCode - 40. 组合总和 II ^]
 
-Given a collection of candidate numbers (`candidates`) and a target number (`target`), find all unique combinations in `candidates` where the candidate numbers sums to `target`.
+给定一个候选人编号的集合 `candidates` 和一个目标数 `target` ，找出 `candidates` 中所有可以使数字和为 `target` 的组合。
 
-Each number in `candidates` may only be used once in the combination.
+`candidates` 中的每个数字在每个组合中只能使用 *一次* 。
 
-*Note:*
+**注意：**解集不能包含重复的组合。
 
-* All numbers (including target) will be positive integers.
-* The solution set must not contain duplicate combinations.
+*示例 1:*
 
-.Example 1:
-[source]
-----
-Input: candidates = [10,1,2,7,6,1,5], target = 8,
-A solution set is:
+....
+输入: candidates = [10,1,2,7,6,1,5], target = 8,
+输出:
 [
-  [1, 7],
-  [1, 2, 5],
-  [2, 6],
-  [1, 1, 6]
+  [1,1,6],
+  [1,2,5],
+  [1,7],
+  [2,6]
 ]
-----
+....
 
-.Example 2:
-[source]
-----
-Input: candidates = [2,5,2,1,2], target = 5,
-A solution set is:
+*示例 2:*
+
+....
+输入: candidates = [2,5,2,1,2], target = 5,
+输出:
 [
   [1,2,2],
   [5]
 ]
-----
+....
+
+
+*提示:*
+
+* `+1 <= candidates.length <= 100+`
+* `+1 <= candidates[i] <= 50+`
+* `+1 <= target <= 30+`
+
 
 == 思路分析
 
@@ -80,8 +85,24 @@ include::{sourcedir}/_0040_CombinationSumII_2.java[tag=answer]
 include::{sourcedir}/_0040_CombinationSumII_3.java[tag=answer]
 ----
 --
+
+四刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0040_CombinationSumIi_4.java[tag=answer]
+----
+--
 ====
 
+
+== 思考题
+
+如果一个既存在重复元素，同一个元素又可以重复使用，该怎么做组合总和？
+
+其实，不能这样搞。没办法区分是同一个元素，还是重复的元素。
+
 == 参考资料
 
 . https://leetcode.cn/problems/combination-sum-ii/solutions/407850/zu-he-zong-he-ii-by-leetcode-solution/[40. 组合总和 II - 官方题解^]

docs/images/0040-10.png

@@ -493,6 +493,11 @@ endif::[]
 |{doc_base_url}/0039-combination-sum.adoc[题解]
 |✅ 回溯
 
+|{counter:codes2503}
+|{leetcode_base_url}/combination-sum-ii/[40. 组合总和 II^]
+|{doc_base_url}/0040-combination-sum-ii.adoc[题解]
+|✅ 回溯。注意同层剪枝的技巧。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

docs/images/0040-11.png

@@ -0,0 +1,50 @@
+package com.diguage.algo.leetcode;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+public class _0040_CombinationSumIi_4 {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-09-16 17:31:07
+   */
+  public List<List<Integer>> combinationSum2(int[] candidates, int target) {
+    Arrays.sort(candidates);
+    List<List<Integer>> result = new ArrayList<>();
+    List<Integer> path = new ArrayList<>();
+    backtrack(candidates, target, result, path, 0);
+    return result;
+  }
+
+  private void backtrack(int[] candidates, int target,
+                         List<List<Integer>> result,
+                         List<Integer> path, int index) {
+    if (target == 0) {
+      result.add(new ArrayList<>(path));
+      return;
+    }
+    for (int i = index; i < candidates.length; i++) {
+      int num = candidates[i];
+      // 已经超量了，则直接返回
+      if (num > target) {
+        return;
+      }
+      // 同一层，相同的数字只能选一次。否则，就会产生重复的组合。
+      if (i > index && candidates[i - 1] == candidates[i]) {
+        continue;
+      }
+      path.add(num);
+      backtrack(candidates, target - num, result, path, i + 1);
+      path.removeLast();
+    }
+  }
+
+  // end::answer[]
+  public static void main(String[] args) {
+    new _0040_CombinationSumIi_4()
+      .combinationSum2(new int[]{10, 1, 2, 7, 6, 1, 5}, 8);
+  }
+}