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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"].
直观思路,判断nums[i]- nums[i-1]是否为1,是就继续,不是就判定范围,存入ArrayList.
Java code:
public class Solution { public List<String> summaryRanges(int[] nums) { List<String> result = new ArrayList<String>(); if(nums == null || nums.length == 0){ return result; } int left = nums[0]; int right = 0; for(int i = 1; i < nums.length; i++){ if(nums[i] - nums[i-1] != 1){ right = nums[i-1]; if(left == right){ result.add("" + left); }else { result.add(left + "->"+right); } left = nums[i]; } } right = nums[nums.length-1]; if(left == right){ result.add("" + left); }else { result.add(left + "->"+right); } return result; } }
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原文地址:http://www.cnblogs.com/anne-vista/p/4868765.html