86 lines
2.2 KiB
Java
86 lines
2.2 KiB
Java
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//给定一个数组 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。
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//
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// candidates 中的每个数字在每个组合中只能使用一次。
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//
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// 注意：解集不能包含重复的组合。
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//
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//
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//
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// 示例 1:
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//
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//
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//输入: candidates = [10,1,2,7,6,1,5], target = 8,
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//输出:
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//[
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//[1,1,6],
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//[1,2,5],
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//[1,7],
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//[2,6]
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//]
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//
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// 示例 2:
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//
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//
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//输入: candidates = [2,5,2,1,2], target = 5,
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//输出:
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//[
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//[1,2,2],
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//[5]
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//]
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//
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//
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//
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// 提示:
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//
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//
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// 1 <= candidates.length <= 100
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// 1 <= candidates[i] <= 50
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// 1 <= target <= 30
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//
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// Related Topics 数组 回溯 👍 668 👎 0
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package leetcode.editor.cn;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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//40:组合总和 II
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class CombinationSumIi {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new CombinationSumIi().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public List<List<Integer>> combinationSum2(int[] candidates, int target) {
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Arrays.sort(candidates);
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List<List<Integer>> result = new ArrayList<>();
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dfs(candidates, target, 0, result, new ArrayList<>());
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return result;
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}
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private void dfs(int[] candidates, int target, int index, List<List<Integer>> result,
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List<Integer> list) {
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if (target == 0) {
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result.add(new ArrayList<>(list));
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return;
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}
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if (index >= candidates.length || target < 0) {
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return;
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}
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for (int i = index; i < candidates.length; i++) {
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if (i - 1 < candidates.length && i > 0 && i > index && candidates[i] == candidates[i - 1]) {
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continue;
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}
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list.add(candidates[i]);
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dfs(candidates, target - candidates[i], i + 1, result, list);
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list.remove(list.size() - 1);
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}
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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