Combination Sum II - Leetcode

最新推荐文章于 2021-02-13 16:07:11 发布

AngelQian2

最新推荐文章于 2021-02-13 16:07:11 发布

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分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/Q_wing/article/details/43864837

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本文介绍了解决组合总和问题的一种高效算法。通过递归深度优先搜索(DFS)并结合排序和去重策略，实现了在不重复使用元素的情况下找出所有可能的组合，使得这些组合中元素的和等于目标值。

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public class Solution {
    public List<List<Integer>> combinationSum2(int[] num, int target) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> path  = new ArrayList<>();
        Arrays.sort(num);
        DFS(result, path, num, target, 0);
        return result;
    }
    private void DFS(List<List<Integer>> result, List<Integer> path, int[] pool, int gap, int start){
        if(gap == 0){
            ArrayList<Integer> l = new ArrayList<>(path);
            if(!result.contains(l))
                result.add(l);
            return;
        }
        for(int i=start; i<pool.length; i++){
            if(gap < pool[i])
               return;
            path.add(pool[i]);
            DFS(result,path,pool,gap-pool[i],i+1);
            path.remove(path.size()-1);
        }
    }
}

2）

public class Solution {
    public List<List<Integer>> combinationSum2(int[] num, int target) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> path  = new ArrayList<>();
        Arrays.sort(num);
        DFS(result, path, num, target, 0);
        return result;
    }
    private void DFS(List<List<Integer>> result, List<Integer> path, int[] pool, int gap, int start){
        if(gap == 0){
            ArrayList<Integer> l = new ArrayList<>(path);
           // if(!result.contains(l))
                result.add(l);
            return;
        }
        int previous = -1;
        for(int i=start; i<pool.length; i++){
            if(gap < pool[i])
               return;
            <span style="background-color: rgb(255, 204, 204);">if(previous == pool[i])
               continue;</span>
               
            previous = pool[i];   
            
            path.add(pool[i]);
            DFS(result,path,pool,gap-pool[i],i+1);
            path.remove(path.size()-1);
        }
    }
}

分析：求所有可能的结果集。第一种解法思路跟之前一样，唯一不同的是，1）每次从下一个元素开始遍历 2）对于结果集验证是否有重复的

第二个解法，效率提高很多，这里用到一个变量记录之前的值

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums toT.

Each number in C may only be used once in the combination.

Note:

All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.