leetCode 40.Combination Sum II(组合总和II) 解题思路和方法
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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:此题和Combination Sum类同,而且很多代码都可以复用,仅仅改变几处不大的变化和增加了一个去处重复项的函数。
具体代码如下:
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>>();//去除重复 for(int i = 0; i < list.size();i++){ if(set.add(list.get(i))){//没有重复 newList.add(list.get(i));//添加新的列表 } } return newList;} public List<List<Integer>> combinationSum1(int[] a, int t) { List<List<Integer>> list = new ArrayList<List<Integer>>(); Arrays.sort(a);//数组排序 //各种特殊情况 if(a.length == 0 || a[0] > t) return list; int len = 0; boolean isAdd = false;//控制与t相等的数只添加一次 for(int i = 0; i< a.length;i++){ if(a[i] == t){ if(!isAdd){//如果未添加 List<Integer> al = new ArrayList<Integer>(); al.add(t); list.add(al); isAdd = true;//标记已添加 } }else if(a[i] < t){//只要比target小的值,大的值肯定不满足,排除 a[len++] = a[i];//新数组 } } //只存在a[0] < target 或 a[0] > target if(a[0] > t)//肯定已没有符合要求的组合 return list; //a[0] < target for(int i = 0; i < len; i++){//循环搜索符合要求的数字组合 int[] b = Arrays.copyOfRange(a, i+1, len);//不含>=t数据的新数组 if(a[i] > t)//如果a[i],肯定以后的数据已不符合,返回 break; //相等于已经有了一个值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){//里面有符合要求的数据 for(int j = 0; j < newList.size();j++){ newList.get(j).add(a[i]);//新返回的各个值添加a[0] Collections.sort(newList.get(j));//排序 list.add(newList.get(j));//最终添加 } } } return list; }}
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