leetCode 40.Combination Sum II(組合總和II) 解題思路和方法
阿新 • • 發佈:2019-01-29
Combination Sum II
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.
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.
For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
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.
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.
For example, given candidate set 10,1,2,7,6,1,5 and target 8,
A solution set is:
[1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
思路:此題和Combination Sum類同,而且很多程式碼都可以複用,僅僅改變幾處不大的變化和增加了一個去處重複項的函式。
具體程式碼如下:
public class Solution { public List<List<Integer>> combinationSum2(int[] a, int t) { List<List<Integer>> list = new ArrayList<List<Integer>>(); List<List<Integer>> newList = new ArrayList<List<Integer>>(); list = combinationSum1(a,t); Set<List<Integer>> set = new HashSet<List<Integer>>();//去除重複 for(int i = 0; i < list.size();i++){ if(set.add(list.get(i))){//沒有重複 newList.add(list.get(i));//新增新的列表 } } return newList; } public List<List<Integer>> combinationSum1(int[] a, int t) { List<List<Integer>> list = new ArrayList<List<Integer>>(); Arrays.sort(a);//陣列排序 //各種特殊情況 if(a.length == 0 || a[0] > t) return list; int len = 0; boolean isAdd = false;//控制與t相等的數只新增一次 for(int i = 0; i< a.length;i++){ if(a[i] == t){ if(!isAdd){//如果未新增 List<Integer> al = new ArrayList<Integer>(); al.add(t); list.add(al); isAdd = true;//標記已新增 } }else if(a[i] < t){//只要比target小的值,大的值肯定不滿足,排除 a[len++] = a[i];//新陣列 } } //只存在a[0] < target 或 a[0] > target if(a[0] > t)//肯定已沒有符合要求的組合 return list; //a[0] < target for(int i = 0; i < len; i++){//迴圈搜尋符合要求的數字組合 int[] b = Arrays.copyOfRange(a, i+1, len);//不含>=t資料的新陣列 if(a[i] > t)//如果a[i],肯定以後的資料已不符合,返回 break; //相等於已經有了一個值a[0]了 List<List<Integer>> newList = new ArrayList<List<Integer>>(); if(i < len -1) newList = combinationSum1(b,t-a[i]); if(newList.size() > 0){//裡面有符合要求的資料 for(int j = 0; j < newList.size();j++){ newList.get(j).add(a[i]);//新返回的各個值新增a[0] Collections.sort(newList.get(j));//排序 list.add(newList.get(j));//最終新增 } } } return list; } }