LeetCode - Easy - 228. Summary Ranges

Topic

• Array
• Two Pointers

Description

https://leetcode.com/problems/summary-ranges/

You are given a sorted unique integer array nums.

Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums.

Each range [a,b] in the list should be output as:

• "a->b" if a != b
• "a" if a == b

Example 1:

Input: nums = [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Explanation: The ranges are:
[0,2] --> "0->2"
[4,5] --> "4->5"
[7,7] --> "7"

Example 2:

Input: nums = [0,2,3,4,6,8,9]
Output: ["0","2->4","6","8->9"]
Explanation: The ranges are:
[0,0] --> "0"
[2,4] --> "2->4"
[6,6] --> "6"
[8,9] --> "8->9"

Example 3:

Input: nums = []
Output: []

Example 4:

Input: nums = [-1]
Output: ["-1"]

Example 5:

Input: nums = [0]
Output: ["0"]

Constraints:

• 0 <= nums.length <= 20
• -2³¹ <= nums[i] <= 2³¹ - 1
• All the values of nums are unique.
• nums is sorted in ascending order.

Analysis

略

Submission

public class SummaryRanges {
     
	
	//方法一：我写的
	public List<String> summaryRanges1(int[] nums) {
     
		List<String> result = new ArrayList<>();
		if (nums == null || nums.length == 0)
			return result;

		int left = nums[0], right = nums[0];

		for (int i = 1; i < nums.length; i++) {
     
			if (right + 1 == nums[i]) {
     
				right++;
			} else {
     
				result.add(left == right ? left + "" : left + "->" + right);
				left = right = nums[i];
			}
		}
		result.add(left == right ? left + "" : left + "->" + right);
		return result;
	}

	//方法二：别人写的，双指针
	public List<String> summaryRanges2(int[] nums) {
     
		List<String> list = new ArrayList<>();
		for (int i = 0; i < nums.length; i++) {
     
			int a = nums[i];
			while (i + 1 < nums.length && nums[i + 1] - nums[i] == 1) 
				i++;
			list.add(a == nums[i] ? a + "" : a + "->" + nums[i]);
		}
		return list;
	}
}

Test

import static org.junit.Assert.*;

import org.hamcrest.collection.IsEmptyCollection;

import static org.hamcrest.MatcherAssert.assertThat;
import static org.hamcrest.collection.IsIterableContainingInOrder.contains;

import org.junit.Test;

public class SummaryRangesTest {
     

	@Test
	public void test() {
     
		SummaryRanges obj = new SummaryRanges();

		assertThat(obj.summaryRanges1(new int[] {
     0, 1, 2, 4, 5, 7}), //
				contains("0->2", "4->5", "7"));
		assertThat(obj.summaryRanges1(new int[] {
     0, 2, 3, 4, 6, 8, 9}), // 
				contains("0","2->4","6","8->9"));
		assertThat(obj.summaryRanges1(new int[] {
     }), IsEmptyCollection.empty());
		
		assertThat(obj.summaryRanges1(new int[] {
     -1}), contains("-1"));
		assertThat(obj.summaryRanges1(new int[] {
     0}), contains("0"));
		
		
		assertThat(obj.summaryRanges2(new int[] {
     0, 1, 2, 4, 5, 7}), //
				contains("0->2", "4->5", "7"));
		assertThat(obj.summaryRanges2(new int[] {
     0, 2, 3, 4, 6, 8, 9}), // 
				contains("0","2->4","6","8->9"));
		assertThat(obj.summaryRanges2(new int[] {
     }), IsEmptyCollection.empty());
		
		assertThat(obj.summaryRanges2(new int[] {
     -1}), contains("-1"));
		assertThat(obj.summaryRanges2(new int[] {
     0}), contains("0"));
		
	}
}