228.Summary Ranges
Difficulty: Medium
Given a sorted integer array without duplicates, return the summary of its ranges.
For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].
1. 解法,最简单,遍历判断。
public class Solution { public List<String> summaryRanges(int[] nums) { int length = nums.length; List<String> result = new ArrayList<String>(); if (length == 0) { return result; } int start = nums[0]; int theoreticalValue = start + 1; for (int index = 0; index < length; index ++) { if (index == length - 1) { if (nums[index] == (theoreticalValue - 1) && start != nums[index]) { String item = start + "->" + nums[index]; result.add(item); } else { String item = "" + nums[index]; result.add(item); } } else if(nums[index + 1] != theoreticalValue) { if (start == nums[index]) { String item = "" + nums[index]; result.add(item); start = nums[index + 1]; theoreticalValue = start; } else { String item = start + "->" + nums[index]; result.add(item); start = nums[index + 1]; theoreticalValue = start; } } theoreticalValue ++; } return result; } }
虽然这个算法简单,当是并没有很好的效率,至打败了5.3%的java提交,这个题目本身不难,但是基本分成三类 ,好像是在第二类?好吧,其实用Python有更加就简单的方法。