228. Summary Ranges

Given a sorted integer array without duplicates, return the summary of its ranges.Example 1:Input: [0,1,2,4,5,7]Output: [“0->2”,“4->5”,“7”]Explanation: 0,1,2 form a continuous range; 4,5 form a continuous range.Example 2:Input: [0,2,3,4,6,8,9]Output: [“0”,“2->4”,“6”,“8->9”]Explanation: 2,3,4 form a continuous range; 8,9 form a continuous range.难度：medium题目：给定排序且无重复元素的整数数组，返回其连续元素的范围。思路：用以变量记录连续元素的开始。Runtime: 5 ms, faster than 7.57% of Java online submissions for Summary Ranges.Memory Usage: 37.5 MB, less than 5.02% of Java online submissions for Summary Ranges.class Solution { public List<String> summaryRanges(int[] nums) { List<String> result = new ArrayList<>(); if (null == nums || nums.length < 1) { return result; } for (int i = 1, start = 0; i <= nums.length; i++) { if (i == nums.length || nums[i] - nums[i - 1] != 1) { result.add((i - 1 == start) ? String.format("%s", nums[start]) : String.format("%s->%s", nums[start], nums[i - 1])); start = i; } } return result; }}