LeetCode | 易学教程

Topic

Array
Two Pointers

Description

Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Notice that the solution set must not contain duplicate triplets.

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

Example 2:

Input: nums = []
Output: []

Example 3:

Input: nums = [0]
Output: []

Constraints:

0 <= nums.length <= 3000
-10⁵ <= nums[i] <= 10⁵

Analysis

方法一：

首先将数组从左到右升序排序。
接下来动用三个指针，用一指针a先固定一元素，然后指针b初始位置为a的右一位，指针c初始位置数组尾元素，
b、c两指针向中间夹逼寻找出这两指针所指元素之和等于a指针所指元素 * -1，期间忽略重复元素，寻找到则将结果添加至List。
当b、c两指针相遇，指针a向右移动一位，跳到步骤2.，直到指针a移动到数组倒数第二位。

方法二：与方法一类似，用到Set去重。

方法三：在方法一的基础上，进一步完善：

方法体内第一句后添加防御性代码if (nums.length < 3) return res;。
for循环中第一语句前插入一句if (nums[i] > 0) break;。这是因为数组已经是从左到右升序的，如果第一个数大于0，就不可能在接下来的元素中寻找出两元素之和小于0的。
将if (i == 0 || (i > 0 && num[i] != num[i-1]))中的i > 0是不必要的，将其移除。

Submission

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Set;

public class ThreeSum {

	// 方法一：
	public List<List<Integer>> threeSum1(int[] num) {
		Arrays.sort(num);
		List<List<Integer>> res = new LinkedList<>();
		for (int i = 0; i < num.length - 2; i++) {
			if (i == 0 || (i > 0 && num[i] != num[i - 1])) {
				int lo = i + 1, hi = num.length - 1, sum = 0 - num[i];
				while (lo < hi) {
					if (num[lo] + num[hi] == sum) {
						res.add(Arrays.asList(num[i], num[lo], num[hi]));
						while (lo < hi && num[lo] == num[lo + 1])
							lo++;
						while (lo < hi && num[hi] == num[hi - 1])
							hi--;
						lo++;
						hi--;
					} else if (num[lo] + num[hi] < sum)
						lo++;
					else
						hi--;
				}
			}
		}
		return res;
	}

	// 方法二：与方法一类似，用到Set去重
	public List<List<Integer>> threeSum2(int[] nums) {
		Set<List<Integer>> res = new HashSet<>();
		if (nums.length == 0)
			return new ArrayList<>(res);
		Arrays.sort(nums);
		for (int i = 0; i < nums.length - 2; i++) {
			int j = i + 1;
			int k = nums.length - 1;
			while (j < k) {
				int sum = nums[i] + nums[j] + nums[k];
				if (sum == 0)
					res.add(Arrays.asList(nums[i], nums[j++], nums[k--]));
				else if (sum > 0)
					k--;
				else if (sum < 0)
					j++;
			}

		}
		return new ArrayList<>(res);
	}

	//方法三：在方法一的基础上做进一步改善
	public List<List<Integer>> threeSum3(int[] nums) {
		List<List<Integer>> res = new ArrayList<>();
		if (nums.length < 3)
			return res;
		Arrays.sort(nums);
		for (int i = 0; i < nums.length - 2; i++) {
			if (nums[i] > 0)
				break;
			if (i == 0 || nums[i] != nums[i - 1]) {
				int lo = i + 1, hi = nums.length - 1, sum = 0 - nums[i];
				while (lo < hi) {
					if (nums[lo] + nums[hi] == sum) {
						res.add(Arrays.asList(nums[i], nums[lo], nums[hi]));
						while (lo < hi && nums[lo] == nums[lo + 1])
							lo++;
						while (lo < hi && nums[hi] == nums[hi - 1])
							hi--;
						lo++;
						hi--;
					} else if (nums[lo] + nums[hi] < sum)
						lo++;
					else
						hi--;
				}
			}
		}
		return res;
	}
}

Test

import static org.junit.Assert.*;
import static org.hamcrest.MatcherAssert.assertThat;
import static org.hamcrest.collection.IsIterableContainingInAnyOrder.containsInAnyOrder;

import java.util.Arrays;

import org.hamcrest.collection.IsEmptyCollection;
import org.junit.Test;

public class ThreeSumTest {

	@Test
	@SuppressWarnings("unchecked")
	public void test() {
		ThreeSum obj = new ThreeSum();

		assertThat(obj.threeSum1(new int[] { -1, 0, 1, 2, -1, -4 }),
				containsInAnyOrder(Arrays.asList(-1, -1, 2), Arrays.asList(-1, 0, 1)));
		assertThat(obj.threeSum1(new int[] {}), IsEmptyCollection.empty());
		assertThat(obj.threeSum1(new int[] { 0 }), IsEmptyCollection.empty());

		assertThat(obj.threeSum2(new int[] { -1, 0, 1, 2, -1, -4 }),
				containsInAnyOrder(Arrays.asList(-1, -1, 2), Arrays.asList(-1, 0, 1)));
		assertThat(obj.threeSum2(new int[] {}), IsEmptyCollection.empty());
		assertThat(obj.threeSum2(new int[] { 0 }), IsEmptyCollection.empty());

		assertThat(obj.threeSum3(new int[] { -1, 0, 1, 2, -1, -4 }),
				containsInAnyOrder(Arrays.asList(-1, -1, 2), Arrays.asList(-1, 0, 1)));
		assertThat(obj.threeSum3(new int[] {}), IsEmptyCollection.empty());
		assertThat(obj.threeSum3(new int[] { 0 }), IsEmptyCollection.empty());
	}
}

来源：oschina

链接：https://my.oschina.net/jallenkwong/blog/4872541

标签

java

ASList