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| 1 | +package com.interview.array; |
| 2 | + |
| 3 | +import java.util.ArrayList; |
| 4 | +import java.util.Collections; |
| 5 | +import java.util.List; |
| 6 | + |
| 7 | +/** |
| 8 | + * Date 10/19/2016 |
| 9 | + * @author Tushar Roy |
| 10 | + * |
| 11 | + * Given a sorted integer array without duplicates, return the summary of its ranges. |
| 12 | + * For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"]. |
| 13 | + * |
| 14 | + * Solution - |
| 15 | + * Just check if num[i] + 1 != num[i + 1]. If its not equal means you need to add previous range to result |
| 16 | + * and start a new range. |
| 17 | + * |
| 18 | + * Time complexity O(n) |
| 19 | + * |
| 20 | + * https://leetcode.com/problems/summary-ranges/ |
| 21 | + */ |
| 22 | +public class SummaryRanges { |
| 23 | + public List<String> summaryRanges(int[] nums) { |
| 24 | + if (nums.length == 0) { |
| 25 | + return Collections.EMPTY_LIST; |
| 26 | + } |
| 27 | + if (nums.length == 1) { |
| 28 | + return Collections.singletonList(String.valueOf(nums[0])); |
| 29 | + } |
| 30 | + int start = 0; |
| 31 | + List<String> result = new ArrayList<>(); |
| 32 | + for (int i = 0; i < nums.length - 1; i++) { |
| 33 | + if ((nums[i] + 1) != nums[i + 1]) { |
| 34 | + result.add(makeRange(nums[start], nums[i])); |
| 35 | + start = i + 1; |
| 36 | + } |
| 37 | + } |
| 38 | + if ((nums[nums.length - 2] + 1) != nums[nums.length - 1]) { |
| 39 | + start = nums.length - 1; |
| 40 | + } |
| 41 | + result.add(makeRange(nums[start], nums[nums.length - 1])); |
| 42 | + return result; |
| 43 | + } |
| 44 | + |
| 45 | + private String makeRange(int a, int b) { |
| 46 | + if (a == b) { |
| 47 | + return String.valueOf(a); |
| 48 | + } |
| 49 | + return a + "->" + b; |
| 50 | + } |
| 51 | +} |
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