40. 组合总和 II

输入: candidates = [10,1,2,7,6,1,5], target = 8,
所求解集为:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

输入: candidates = [2,5,2,1,2], target = 5,
所求解集为:
[
  [1,2,2],
  [5]
]

/*
 * @lc app=leetcode id=40 lang=javascript
 *
 * [40] Combination Sum II
 *
 * https://leetcode.com/problems/combination-sum-ii/description/
 *
 * algorithms
 * Medium (40.31%)
 * Total Accepted:    212.8K
 * Total Submissions: 519K
 * Testcase Example:  '[10,1,2,7,6,1,5]\n8'
 *
 * 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]
 * ]
 *
 *
 */
function backtrack(list, tempList, nums, remain, start) {
    if (remain < 0) return;
    else if (remain === 0) return list.push([...tempList]);
    for (let i = start; i < nums.length; i++) {
      // 和39.combination-sum 的其中一个区别就是这道题candidates可能有重复
      // 代码表示就是下面这一行
      if(i > start && nums[i] == nums[i-1]) continue; // skip duplicates
      tempList.push(nums[i]);
      backtrack(list, tempList, nums, remain - nums[i], i + 1); // i + 1代表不可以重复利用， i 代表数字可以重复使用
      tempList.pop();
    }
  }
/**
 * @param {number[]} candidates
 * @param {number} target
 * @return {number[][]}
 */
var combinationSum2 = function(candidates, target) {
    const list = [];
    backtrack(list, [], candidates.sort((a, b) => a - b), target, 0);
    return list;
};

class Solution:
    def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
        """
        与39题的区别是不能重用元素，而元素可能有重复；
        不能重用好解决，回溯的index往下一个就行；
        元素可能有重复，就让结果的去重麻烦一些；
        """
        size = len(candidates)
        if size == 0:
            return []

        # 还是先排序，主要是方便去重
        candidates.sort()

        path = []
        res = []
        self._find_path(candidates, path, res, target, 0, size)

        return res

    def _find_path(self, candidates, path, res, target, begin, size):
        if target == 0:
            res.append(path.copy())
        else:
            for i in range(begin, size):
                left_num = target - candidates[i]
                if left_num < 0:
                    break
                # 如果存在重复的元素，前一个元素已经遍历了后一个元素与之后元素组合的所有可能
                if i > begin and candidates[i] == candidates[i-1]:
                    continue
                path.append(candidates[i])
                # 开始的 index 往后移了一格
                self._find_path(candidates, path, res, left_num, i+1, size)
                path.pop()