【力扣】40. 组合总和 II

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。

candidates 中的每个数字在每个组合中只能使用 一次 。

注意：解集不能包含重复的组合。 

示例 1:

输入: candidates = [10,1,2,7,6,1,5], target = 8,
输出:
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]

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

题解：

回溯法：

func combinationSum2(candidates []int, target int) [][]int {
	var path []int
	var result [][]int
	var backtracking func(candidates []int, target, sum, startIndex int)
	backtracking = func(candidates []int, target, sum, startIndex int) {
		// 终止条件
		if sum > target {
			return
		}
		if sum == target {
			tmp := make([]int, len(path))
			copy(tmp, path)
			result = append(result, tmp)
			return
		}
		for i := startIndex; i < len(candidates); i++ {
			// 剪枝优化
			if candidates[i] > target {
				break
			}
			// 去重复
			if i > startIndex && candidates[i] == candidates[i-1] {
				continue
			}
			path = append(path, candidates[i])
			sum += candidates[i]
			backtracking(candidates, target, sum, i+1)
			// 回溯
			sum -= candidates[i]
			path = path[:len(path)-1]
		}
	}
	sort.Ints(candidates)
	backtracking(candidates, target, 0, 0)
	return result
}