leecode---39---array, dfs, backtracking method---seek all combinations as target, no repetition of numbers

Given an array and target value, to find a few numbers and the target value.

In dfs, the input parameters are: total result, current digital record, current digital sum, array, array subscript, target value.

If the current sum exceeds target, exit dfs

If the current sum == target, record the result

 

 

---

https://leetcode.com/problems/combination-sum/description/

 

 

meaning of the title

Given an array and target, find all combinations, in which the numbers can be reused, and there are no repeated numbers in the array

 

analyze

http://www.zhimengzhe.com/bianchengjiaocheng/Javabiancheng/257227.html Summary of backtracking

 

 

 

code

class Solution {

    public List<List<Integer>> combinationSum(int[] candidates, int target) {

        List<List<Integer>> result = new ArrayList<List<Integer>>();

        List<Integer> temp = new ArrayList<Integer>();

        if(candidates == null || candidates.length <= 0 || target < 0) return result;

        dfs(result,temp,candidates,0,0,target);

        return result;

    }

    

    public void dfs(List<List<Integer>> result,List<Integer> temp,int[] nums,int pos,int curSum,int target){

        //1. Exit the loop directly

        if (curSum > target) return;

        

        //2. Result situation

        if (curSum == target) {

            result.add(new ArrayList<Integer>(temp));

        }

        

        //3. Ordinary case --- selection condition: because the number can be used repeatedly without repetition, it can be used from pos to the end

        //4. Whether to backtrack---because the distance is recorded, the previous choice needs to be eliminated when making the next choice.

        for (int i=pos; i < nums.length; i++) {

            temp.add(nums[i]);

            curSum+=nums[i];

            dfs(result,temp,nums,i,curSum,target);

            temp.remove(temp.size() - 1);

            curSum = nums[i];

        }

    }

}