LeetCode刷题day040 (Jieky)

LeetCode第40题 Combination Sum II

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

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

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

Example 1:
Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:
Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]
*/
import java.util.*;
public class CombinationSumII{
     
	public static void main(String[] args){
     

		Scanner sc = new Scanner(System.in); 
		
		System.out.println("请输入一串整数，并用空格隔开，以回车结束");
		// 一次读一行：https://www.cnblogs.com/liyao0312/p/11202393.html
		String[] str = sc.nextLine().split(" "); 
		int nums[]=new int[str.length];
		for(int i=0;i<nums.length;i++){
     
			// 字符串转Int
			nums[i]=Integer.parseInt(str[i]);
		}
		
		// Integer[] nums = temp.toArray(new Integer[temp.size()]);
		// int[] nums = temp.stream().mapToInt(Integer::valueOf).toArray();
		
		System.out.println("输入目标：");
		int target = sc.nextInt();
		sc.close();
		
		CombinationSumII cs = new CombinationSumII();
		List<List<Integer>> result = cs.combinationSum(nums, target);
		System.out.println(result);
		
	}
	
	// 穷举的办法
	public List<List<Integer>> combinationSum(int[] nums,int target){
     
		List<List<Integer>> list = new ArrayList<>();
		backtrack(list,nums,target,new ArrayList<Integer>(),0);
		// return list;
		return removeDuplicate(list);
	}

	private void backtrack(List<List<Integer>> list,int[] nums,int remaind,List<Integer> tempList,int start){
     
		if (remaind < 0) {
     
			return;
		}else if(remaind == 0) {
     
			list.add(new ArrayList<>(tempList));
		}else{
     
			for (int i=start;i<nums.length;i++){
     
				tempList.add(nums[i]);
				// jian start=i -> start=i+1
				backtrack(list,nums,remaind-nums[i],tempList,i+1);
				// backtrack运行完只有两种可能：remaind==0 or ramaind < 0
				tempList.remove(tempList.size()-1);
			}
		}
	}
	
	private List<List<Integer>> removeDuplicate(List<List<Integer>> list) {
     
		Map<String, String> ans = new HashMap<String, String>();
		for (int i = 0; i < list.size(); i++) {
     
			List<Integer> l = list.get(i);
			// 对list容器排序
			Collections.sort(l);
			String key = "";
			for (int j = 0; j < l.size() - 1; j++) {
     
				// 将数值拼接成字符串
				key = key + l.get(j) + ",";
			}
			key = key + l.get(l.size() - 1);
			// 若key存在，则覆盖之前的value
			ans.put(key, "");
		}
		List<List<Integer>> ans_list = new ArrayList<List<Integer>>();
		// 遍历去重后的map的KeySet
		for (String k : ans.keySet()) {
     
			String[] l = k.split(",");
			// 遍历list，将字符转化为整数：Integer.parseInt
			List<Integer> temp = new ArrayList<Integer>();
			for (int i = 0; i < l.length; i++) {
     
				int c = Integer.parseInt(l[i]);
				temp.add(c);
			}
			ans_list.add(temp);
		}
		return ans_list;
	}
}