题目地址： Combination Sum II 描述： 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. Each number in candidates may only be used once in the combination. Note: All numbers (including target) will be positive integers. The solution set must not contain duplicate combinations. Example 1: Input: candidates = [10,1,2,7,6,1,5], target = 8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ] Example 2: Input: candidates = [2,5,2,1,2], target = 5, A solution set is: [ [1,2,2], [5] ] 大致意思：给你一个数组（都是正数）和目标数，找出加一起等于目标数的所有的组合，要求这些组合不能重复。
import java.util.*; public class CombinationSumII { public static ArrayList<ArrayList<Integer>> combinationSumII(int[] candidates, int target) { ArrayList<Integer> list = new ArrayList<>(); ArrayList<ArrayList<Integer>> res = new ArrayList<>(); Arrays.sort(candidates); backTracking(candidates, target, list, res, 0); return res; } //1122456 private static void backTracking(int[] candidates, int target, ArrayList<Integer> list, ArrayList<ArrayList<Integer>> res, int count) { if( target <= 0){ return; } int length = candidates.length; for (int i = count; i < length; i++) { //过滤相同元素，此处之所以加上i>count一方面防止i=0报错，另一方面是防止过滤掉类似于1124,116这种有相等值的元素，如果不理解可以去掉跑一遍。 if (i > count && candidates[i]==candidates[i-1]) continue; list.add(candidates[i]); backTracking(candidates, target - candidates[i], list, res, i+1); if (target == candidates[i]) { res.add(new ArrayList<>(list)); } list.remove(list.size() - 1); } } public static void main(String[] args) { int[] candidates = {1, 6, 2, 5, 3,1,2,4}; Arrays.sort(candidates); ArrayList<ArrayList<Integer>> arrayLists = combinationSumII(candidates, 8); System.out.println(arrayLists); } }