[LintCode/LeetCode] Combination Sum I & II

ThreeWords 发布于2019-08-14 15:23 / 2674人阅读

摘要：和唯一的不同是组合中不能存在重复的元素，因此，在递归时将初始位即可。

Combination Sum I Problem

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

The same repeated number may be chosen from C unlimited number of times.

Notice

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.

Example

given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]

Note

基本思路与Combinations一致，递归模板详见：https://segmentfault.com/a/11...
有两个点需要注意：在组合中的数必须是升序排列，所以在调用dfs函数之前要先排序；另外，由于组合里允许有重复数，dfs调用自身时，初始位start(=i)的位置不变，依然从i开始，只需将target减小num[i]即可。

Solution Recursion 26ms

public class Solution {
    List> res = new ArrayList<>();
    public List> combinationSum(int[] candidates, int target) {
        Arrays.sort(candidates);
        helper(candidates, 0, target, new ArrayList());
        return res;
    }
    public void helper(int[] c, int start, int t, List pre) {
        if (t < 0) return;
        if (t == 0) res.add(pre);
        for (int i = start; i < c.length; i++) {
            List cur = new ArrayList(pre);
            cur.add(c[i]);
            helper(c, i, t-c[i], cur);
        }
    }
}

Combination Sum II Problem

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

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

Notice

Example

Given candidate set [10,1,6,7,2,1,5] and target 8,

A solution set is:

[
[1,7],
[1,2,5],
[2,6],
[1,1,6]
]

Note

和Combination Sum I唯一的不同是组合中不能存在重复的元素，因此，在dfs递归时将初始位+1即可。

Solution

public class Solution {
    List> res = new ArrayList>();
    public List> combinationSum2(int[] num, int target) {
        Arrays.sort(num);
        helper(num, 0, target, new ArrayList());
        return res;
    }
    public void helper(int[] num, int start, int target, List pre) {
        if (target < 0) return;
        if (target == 0) {
            res.add(pre);
            return;
        }
        for (int i = start; i < num.length; i++) {
            if (i > start && num[i] == num[i-1]) continue;
            List cur = new ArrayList (pre);
            cur.add(num[i]);
            helper(num, i+1, target-num[i], cur);
        }
    }
}

Combination Sum III & IV link: here

混合云超融合服务器 II ii6 ii7 Combination

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