原题链接在这里：https://leetcode.com/problems/missing-ranges/
题目：
Given a sorted integer array nums, where the range of elements are in the inclusive range [lower, upper], return its missing ranges.
Example:
Input: nums =[0, 1, 3, 50, 75], lower = 0 and upper = 99, Output:["2", "4->49", "51->74", "76->99"]
题解：
If nums is null or empty, just add getRange(lower, upper).
Otherwise, first add range between lower and nums[0].
Then add missing range within nums.
Last, add range between nums[nums.length - 1] and upper.
Time Complexity: O(n). n = nums.length.
Space: O(1). regardless res.
AC Java:
1 class Solution { 2 public List<String> findMissingRanges(int[] nums, int lower, int upper) { 3 List<String> res = new ArrayList<>(); 4 if(nums == null || nums.length == 0){ 5 res.add(getRange(lower, upper)); 6 return res; 7 } 8 9 if(lower < nums[0]){ 10 res.add(getRange(lower, nums[0] - 1)); 11 } 12 13 for(int i = 1; i<nums.length; i++){ 14 if(nums[i - 1] != nums[i] && nums[i - 1] + 1 != nums[i]){ 15 res.add(getRange(nums[i - 1] + 1, nums[i] - 1)); 16 } 17 } 18 19 if(nums[nums.length - 1] < upper){ 20 res.add(getRange(nums[nums.length - 1] + 1, upper)); 21 } 22 23 return res; 24 } 25 26 private String getRange(int l, int r){ 27 if(l == r){ 28 return "" + l; 29 } 30 31 return l + "->" + r; 32 } 33 }