LeetCode:4Sum

Given an array S of n integers, are there elements a, b, c, 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:

  • Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, abcd)
  • 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)
4Sum问题和3Sum问题没太大区别。

public class Solution {
        public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        int length = num.length;
        String content = "";
        if(length < 4) return result;
        sort(num,0,length -1);
        for(int i = 0;i<length -3;i++)
        {
            for(int j = i+1;j<length -2;j++)
            {
                int value = target - num[i] - num[j];
                int start = j + 1;
                int end = length -1;
                while(start < end)
                {
                    if(num[start] + num[end] == value)
                    {
                        String temp = "" + num[i] +","+num[j]+","+num[start]+","+num[end]+",";
                        if(!content.contains(temp))
                        {
                            ArrayList<Integer> order = new ArrayList<Integer>();
                            order.add(num[i]);
                            order.add(num[j]);
                            order.add(num[start]);
                            order.add(num[end]);
                            content = content + temp;
                            result.add(order);
                        }
                        start ++;
                    }
                    else if(num[start] + num[end] > value)
                    {
                        end -- ;
                    }
                    else if(num[start] + num[end] < value)
                    {
                        start ++;
                    }
                }
            }
        }
        return result;
       
    }
    public void sort(int[] num,int start,int end)
    {
        if(start<end)
        {
            int p = partition(num,start,end);
            sort(num,start,p-1);
            sort(num,p+1,end);
        }
       
    }
    public int partition(int[] num,int start,int end)
    {
        if(end - start==0)return start;
        int key = num[start];
        while(start<end)
        {
            while(end>start&&num[end]>key)
            {
                end -- ;
            }
            num[start] = num[end];
            while(start<end&&num[start]<=key)
            {
                start ++;
            }
            num[end] = num[start];
        }
        num[start]=key;
        return start;
    }
}

posted on 2014-05-23 11:34  JessiaDing  阅读(130)  评论(0编辑  收藏  举报