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原题链接在这里：https://leetcode.com/problems/missing-ranges/
题目：
Given a sorted integer array where the range of elements are [lower, upper] inclusive, return its missing ranges.
For example, given [0, 1, 3, 50, 75], lower = 0 and upper = 99, return ["2", "4->49", "51->74", "76->99"].
题解：
与Summary Ranges相似.
expected是当前期待的数字, 初始化为lower.
若是expected不等于nums[i]时就说明出现了missing range, 要分类加结果.
看最后一个元素是否是upper, 若不是，还需要加最后一段.
Time Complexity: O(n). Space: O(1). 不考虑res大小.
AC Java:
1 public class Solution { 2 public List<String> findMissingRanges(int[] nums, int lower, int upper) { 3 List<String> res = new ArrayList<String>(); 4 if(nums == null || nums.length == 0){ 5 if(lower == upper){ 6 res.add(String.valueOf(lower)); 7 return res; 8 }else{ 9 res.add(String.valueOf(lower) + "->" + String.valueOf(upper)); 10 return res; 11 } 12 } 13 14 //最开始的expeted number 是 lower 15 int expected = lower; 16 for(int i = 0; i<nums.length; i++){ 17 if(nums[i] != expected){ //若是出现expected 不等于 nums[i], 就该加missing range 到res中了 18 if(expected + 1 == nums[i]){ 19 res.add(String.valueOf(expected)); 20 }else{ 21 res.add(String.valueOf(expected) + "->" + String.valueOf(nums[i]-1)); 22 } 23 } 24 //不论expected 是否等于 nums[i], 都更新expected 到 num[i]+1. 25 expected = nums[i]+1; 26 27 //到了nums最后一个还是没有到达upper 28 if(i == nums.length-1 && nums[i] != upper){ 29 if(nums[i] + 1 == upper){ 30 res.add(String.valueOf(nums[i]+1)); 31 }else{ 32 res.add(String.valueOf(nums[i]+1) + "->" + String.valueOf(upper)); 33 } 34 } 35 } 36 37 return res; 38 } 39 }
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原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/5301190.html
