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LeetCode - 4Sum

时间:2015-12-20 08:10:37      阅读:186      评论:0      收藏:0      [点我收藏+]

题目:

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, a ≤ b ≤ c ≤ d)
    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)

题目:

延续3sum和2sum的思路,不断往下找,注意排除重复的元素。

package sum;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class FourSum {

    public List<List<Integer>> fourSum(int[] nums, int target) {
        int len;
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (nums == null || (len = nums.length) < 4) return res;
        Arrays.sort(nums);
        for (int i = 0; i < len; ++i) {
            List<Integer> subRes = new ArrayList<Integer>();
            subRes.add(nums[i]);
            getOtherThree(res, subRes, nums, i + 1, len, target);
            while (i + 1 < len && nums[i + 1] == nums[i]) ++i;
        }
        
        return res;
    }
    
    private void getOtherThree(List<List<Integer>> res, List<Integer> subRes, int[] nums, int start, int len, int target) {
        for (int i = start; i < len; ++i) {
            List<Integer> copySubRes = new ArrayList<Integer>(subRes);
            copySubRes.add(nums[i]);
            getOtherTwo(res, copySubRes, nums, i + 1, len, target);
            while (i + 1 < len && nums[i + 1] == nums[i]) ++i;
        }
    }
    
    private void getOtherTwo(List<List<Integer>> res, List<Integer> subRes, int[] nums, int start, int len, int target) {
        int sum = 0;
        for (int i : subRes) sum += i;
        int rem = target - sum;
        
        int left = start;
        int right = len - 1;
        while (left < right) {
            if (nums[left] + nums[right] == rem) {
                List<Integer> copySubRes = new ArrayList<Integer>(subRes);
                copySubRes.add(nums[left]);
                copySubRes.add(nums[right]);
                res.add(copySubRes);
                do { ++left; } while (left < len && nums[left - 1] == nums[left]);
                do { --right; } while (right >=0 && nums[right + 1] == nums[right]);
            } else if (nums[left] + nums[right] < rem) {
                ++left;
            } else {
                --right;
            }
        }
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[] nums = { 0,0,4,-2,-3,-2,-2,-3 };
        FourSum f = new FourSum();
        List<List<Integer>> res = f.fourSum(nums, -1);
        for (List<Integer> l : res) {
            for (int i : l)
                System.out.print(i + "\t");
            System.out.println();
        }
    }

}

 

LeetCode - 4Sum

原文:http://www.cnblogs.com/shuaiwhu/p/5060227.html

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