leetcode40. Combination Sum II

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本文介绍了解决LeetCode上的40.组合总和II问题的方法。该问题要求在给定候选数字集合与目标数字的情况下，找出所有唯一组合，使得候选数字之和等于目标数字。文章详细解释了解决方案，并给出了具体的实现代码。

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40. Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C 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.
For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:

[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

解法

在leetcode39. Combination Sum I的基础上改动了两处地方：

backtrack(ret, temp, candidates, remain candidates[i], i + 1); //不包含本次元素
if (i > start && candidates[i] == candidates[i- 1]) continue; // 去除重复的list，如[[1, 7], [1, 7]]，i>start为了避免数组越界，如0-1=-1

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> ret = new ArrayList<>();
        if (candidates == null || candidates.length == 0 || target <= 0) {
            return ret;
        }
        Arrays.sort(candidates);
        backtrack(ret, new ArrayList<Integer>(), candidates, target, 0);
        return ret;
    }
    private void backtrack(List<List<Integer>> ret, List<Integer> temp, int[] candidates, int remain, int start) {
        if (remain == 0) {
            ret.add(new ArrayList<Integer>(temp));
        } else if (remain < 0) {
            return;
        }
        for (int i = start; i < candidates.length; i++) {
            if (i > start && candidates[i] == candidates[i- 1]) continue;
            temp.add(candidates[i]);
            backtrack(ret, temp, candidates, remain - candidates[i], i + 1);
            temp.remove(temp.size() - 1);
        }
    }
}