【Leetcode】combination-sum

题目描述

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.

Note:

• All numbers (including target) will be positive integers.
• Elements in a combination (a 1, a 2, … , a k) must be in non-descending order. (ie, a 1 ≤ a 2 ≤ … ≤ a k).
• The solution set must not contain duplicate combinations.

For example, given candidate set2,3,6,7and target7, 
A solution set is: 
[7]
[2, 2, 3]

题目分析

给定一组候选数字（C，是个set）和目标数字T，找到C中的所有唯一组合，其中候选数字总和为T. 可以从C无限次数中选择相同的重复数。

解题思路

1.首先 从小到大排序（快排）

2.然后回溯法，由于数字set有序，回溯的点：

当加上当前数时，累加的数字之和大于T，因此退一步回到比他小的那个数字上

当加上当前数时，累加的数字之和等于T，当前存放数字序列的集合满足条件，是个解，那再往深的遍历，也没有可用的解了。退到上一步

ac代码

import java.util.ArrayList;
import java.util.List;

public class Solution {
	public ArrayList> combinationSum(int[] candidates, int target) {
		ArrayList> resList = new ArrayList<>();
		if (candidates == null || candidates.length == 0) {
			return resList;
		}
		qsort(candidates, 0, candidates.length-1);

		Integer accu = 0;
		List tempRes = new ArrayList<>();
		dfs(candidates, target, 0, accu, tempRes, resList);
		return resList;
	}

	private static void dfs(int[] candidates, int target, int start, int accu, List tempRes,
			ArrayList> resList) {
		for (int i = start; i < candidates.length; ++i) {
			// 满足条件
			if (accu + candidates[i] == target) {
				tempRes.add(candidates[i]);
				resList.add(new ArrayList<>(tempRes));
				tempRes.remove(tempRes.size() - 1);
				return;
			}
			if (accu + candidates[i] > target) {
				return;
			}
			tempRes.add(candidates[i]);
			dfs(candidates, target, i, accu + candidates[i], tempRes, resList);
			tempRes.remove(tempRes.size() - 1);

		}

	}

	private static void qsort(int[] candidates, int start, int end) {
		if (start >= end) {
			return;
		}
		// 枢轴量
		int temp = candidates[start];
		int left = start;
		int right = end;
		while (left < right) {
			while (right > left && candidates[right] >= temp) {
				--right;
			}
			if (right > left) {
				candidates[left] = candidates[right];
				++left;
			}
			while (right > left && candidates[left] <= temp) {
				++left;
			}
			if (right > left) {
				candidates[right] = candidates[left];
				--right;
			}
		}
		candidates[left] = temp;
		qsort(candidates, start, left - 1);
		qsort(candidates, left + 1, end);

	}

}

经验和教训

1.在深搜过程中 ，

dfs(candidates, target, i , accu + candidates[i], tempRes, resList);

一开始写start传的是i+1，这样候选数字中的每个数字只能使用一次，改为i后，满足了：The same repeated number may be chosen from C unlimited number of times.

2.多写写堆排序