【17】Backtracking algorithm

Idea:

Loop one level at a time, call recursion in each level loop, judge if it meets the requirements, and then return the traceback.

A bit abstract like finding your keys at home:

Start from the first floor of the house (for first floor loop (i=0))----->enter room a------>open the first box [not found] recursion

                                   No more boxes---[Not found] Open the second box <-----Close the first box Backtracking

                                   Exit room a------>enter room b----->open the first box...

Here entering the room and opening the box is like a recursive call, closing the box and closing the door is like backtracking.

Then here the accumulated sum == the goal is to find the key, otherwise we will continue to search one by one.                                                                 

      

Code:

func combinationSum(candidates []int, target int) [][]int {
	
    //定义外部变量
    var result [][]int
	var currentCombination []int

	sort.Ints(candidates) // 首先对数组进行排序

	var backtrack func(startIdx, currentSum int)
	backtrack = func(startIdx, currentSum int) {
        //1）出口
		if currentSum == target {
			// 如果当前和等于目标值，将当前组合添加到结果中
			result = append(result, append([]int{}, currentCombination...))
			return
		}
        //2）开始遍历
		for i := startIdx; i < len(candidates); i++ {
            //1、和大于目标值
			if currentSum+candidates[i] > target {
				break // 如果当前和加上当前元素已经超过目标值，跳出循环
			}
            //2、排除重复项
			if i > startIdx && candidates[i] == candidates[i-1] {
				continue // 跳过重复的元素，避免重复的组合
			}
            // 若 和 小于 目标值 
            //放入这个元素
			currentCombination = append(currentCombination, candidates[i]) // 将当前元素加入当前组合
            //进入递归
			backtrack(i, currentSum+candidates[i])                         // 递归调用，继续查找下一个元素
            //回溯出来的操作：取出这个元素
			currentCombination = currentCombination[:len(currentCombination)-1] // 回溯，移除当前元素
		}
	}

	backtrack(0, 0) // 从第一个元素开始查找

	return result
}