标签：and contain ++ solution code add nat tin sed
题目：
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:
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] ]
题意及分析：这道题比上一题多了一个约束条件，即元素不能重复使用，所以只需要在下一次回溯时将起始点的位置加一即可。具体见代码。
代码：
public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
Arrays.sort(candidates);
List<Integer> array= new ArrayList<>();
List<List<Integer>> list = new ArrayList<>();
int start=0;
int remain=target;
backTracking(list,array,candidates,start,remain);
return list;
}
public void backTracking(List<List<Integer>> list,List<Integer> array,int[] candidates,int start,int remain) {
if(remain<0)
return;
else if(0==remain){ //有解
list.add(new ArrayList<>(array));
}else{
for(int i=start;i<candidates.length;i++){
if(i>start&&candidates[i]==candidates[i-1]) continue; //去重
array.add(candidates[i]);
backTracking(list, array, candidates, i+1, remain-candidates[i]); //这里是i+1
array.remove(array.size()-1);
}
}
return;
}
}
[LeetCode] 40. Combination Sum II Java
标签：and contain ++ solution code add nat tin sed
原文地址：http://www.cnblogs.com/271934Liao/p/6944002.html
