[leetcode]18.4Sum(Java实现)

leetcode地址：https://leetcode.com/problems/4sum/#/description

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: 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]
]

Subscribe to see which companies asked this question.

Java code：

package go.jacob.day619;

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

public class Demo1 {
	/*
	 * accepted java O(n^3) solution based on 3sum
	 * 基于3sum的解法
	 */
	public List> fourSum(int[] nums, int target) {
		List> result = new ArrayList>();
		if (nums == null || nums.length < 4)
			return result;
		Arrays.sort(nums);
		for (int i = 0; i < nums.length - 3; i++) {
			/*
			 * 后面两个if判断可以大大缩短测试时间：185ms到31ms
			 */
			// first candidate too large, search finished
			if (nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target)
				break;
			// first candidate too small
			if (nums[i] + nums[nums.length - 1] + nums[nums.length - 2] + nums[nums.length - 3] < target)
				continue;
			if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) {
				for (int j = i + 1; j < nums.length - 2; j++) {
					// second candidate too large
					if (nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target)
						break; 
					// second candidate too small
					if (nums[i] + nums[j] + nums[nums.length - 1] + nums[nums.length - 2] < target)
						continue; 
					if (j == i + 1 || (j > i + 1 && nums[j] != nums[j - 1])) {
						int sum = target - nums[i] - nums[j], left = j + 1, right = nums.length - 1;
						while (left < right) {
							if (nums[left] + nums[right] > sum)
								right--;
							else if (nums[left] + nums[right] < sum)
								left++;
							else {
								// 找到数对
								ArrayList list = new ArrayList();
								list.add(nums[i]);
								list.add(nums[j]);
								list.add(nums[left]);
								list.add(nums[right]);
								result.add(list);
								left++;
								right--;
							}
							while (left != j + 1 && left < right && nums[left] == nums[left - 1])
								left++;
							while (right != nums.length - 1 && right > left && nums[right] == nums[right + 1])
								right--;
						}
					}
				}
			}
		}
		return result;
	}
}