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【LeetCode 39】组合总和

·287 words·2 mins
WFUing
Author
WFUing
A graduate who loves coding.
Table of Contents

  • 链接：https://leetcode.cn/problems/combination-sum/description/

版本一
#

按照回溯的板子写了一版代码

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        int n = candidates.length;
        dfs(res, candidates, n, 0, target, new ArrayList<Integer>());
        return res;
    }

    public void dfs(List<List<Integer>> res, int[] candidates, int n, int now, int target, ArrayList<Integer> temp) {
        if(now == target) {
            res.add(new ArrayList<>(temp));
            return;
        }
        if(now > target) {
            return;
        }
        for(int i = begin; i < n;i++) {
            temp.add(candidates[i]);
            dfs(res, candidates, n, now+candidates[i], target, temp);
            temp.remove(temp.size()-1);
        }
    }
}

这版本代码会有重复，怎么避免这个问题呢？

版本二
# 

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        int n = candidates.length;
        dfs(res, candidates, 0, n, 0, target, new ArrayList<Integer>());
        return res;
    }

    public void dfs(List<List<Integer>> res, int[] candidates, int begin, int n, int now, int target, ArrayList<Integer> temp) {
        if(now == target) {
            res.add(new ArrayList<>(temp));
            return;
        }
        if(now > target) {
            return;
        }
        for(int i = begin; i < n;i++) {
            temp.add(candidates[i]);
            dfs(res, candidates, i, n, now+candidates[i], target, temp);
            temp.remove(temp.size()-1);
        }
    }
}

版本三 剪枝
# 

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Deque;
import java.util.List;

public class Solution {

    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        int len = candidates.length;
        List<List<Integer>> res = new ArrayList<>();
        if (len == 0) {
            return res;
        }

        // 排序是剪枝的前提
        Arrays.sort(candidates);
        Deque<Integer> path = new ArrayDeque<>();
        dfs(candidates, 0, len, target, path, res);
        return res;
    }

    private void dfs(int[] candidates, int begin, int len, int target, Deque<Integer> path, List<List<Integer>> res) {
        // 由于进入更深层的时候，小于 0 的部分被剪枝，因此递归终止条件值只判断等于 0 的情况
        if (target == 0) {
            res.add(new ArrayList<>(path));
            return;
        }

        for (int i = begin; i < len; i++) {
            // 重点理解这里剪枝，前提是候选数组已经有序，
            if (target - candidates[i] < 0) {
                break;
            }
            
            path.addLast(candidates[i]);
            dfs(candidates, i, len, target - candidates[i], path, res);
            path.removeLast();
        }
    }
}