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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"].
从第一个元素开始,每次保留开始的元素的值,依次扫描后面的元素,如果有连续的就一直往后查找,直到遇到不连续的,然后记录这段范围。直到遍历完所有的元素。代码如下:
For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].
从第一个元素开始,每次保留开始的元素的值,依次扫描后面的元素,如果有连续的就一直往后查找,直到遇到不连续的,然后记录这段范围。直到遍历完所有的元素。代码如下:
public class Solution { public List<String> summaryRanges(int[] nums) { List<String> list = new ArrayList<String>(); if(nums == null || nums.length == 0) return list; for(int i = 0; i < nums.length; i++) { int num = nums[i]; while(i < nums.length - 1 && nums[i] + 1 == nums[i + 1]) { i ++; } if(num != nums[i]) { list.add(num + "->" + nums[i]); } else { list.add(String.valueOf(num)); } } return list; } }
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