leetCode-nSum

Given an array S of n integers, are there elements abc, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

 通用解决办法

public static List<List<Integer>> kSum(int[] nums, int target) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
Arrays.sort(nums);
result = recursionRoutin(nums,0,4,0);
return result;
}

public static List<List<Integer>> recursionRoutin(int[] nums,int begin,int k,int target){
HashSet<List<Integer>> elementSet = new HashSet<List<Integer>>();
List<List<Integer>> result = new ArrayList<List<Integer>>();
List<List<Integer>> subResult = new ArrayList<List<Integer>>();
//Recursion Base
if(k == 2){
int left = begin;
int right = nums.length - 1;
while(left < right){
int sum = nums[left] + nums[right];
if(sum == target){
List<Integer> taplet = new ArrayList<Integer>();
taplet.add(nums[left]);
taplet.add(nums[right]);
//Avoid reduplication
if(!elementSet.contains(taplet)){
result.add(taplet);
elementSet.add(taplet);
}
left ++;
right --;
}else if(sum < target){
left ++;
}else{
right --;
}
}
return result;
}else{
for(int i = begin;i < nums.length - k - 1;i ++){
subResult = recursionRoutin(nums,i + 1,k - 1,target - nums[i]);
//System.out.println(k + " " + subResult);
if (!subResult.isEmpty()) {
for (int j = 0; j < subResult.size(); j++) {
subResult.get(j).add(nums[i]);
result.add(subResult.get(j));
}
}
}
}
return result;
}

  

posted @ 2018-01-06 21:37  知行-zhixing  阅读(516)  评论(0编辑  收藏  举报