M
tags: Array
#### Basic Implementation
- O(n)
- 两个pointer， 每次计较prev和curr之间的部分.
- 然后prev = curr，向前移动一格
- TODO: check the edge case and make sure max/min of int are checked
```
/*
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"].
Tags: Array
Similar Problems: (E) Summary Ranges
*/
/*
Attempt2, Thoughts:
Use two pointer to mark the prev and curr value, then verify the range in between.
matching conditoin: prev +2 >= curr.
That is,
1,...,3
1. When print range: print the missing [x,y]
2. missing x = prev+1, missing y = curr - 1;
3. Make sure prev represents the consecutive integer before missing x.
*/
public class Solution {
public List findMissingRanges(int[] nums, int lower, int upper) {
List rst = new ArrayList();
if (nums == null || nums.length == 0) {//Though, also covered in the for
rst.add(printRange(lower, upper));
return rst;
} else if (lower > upper) {
return rst;
}
int prev = lower - 1;
int curr;
for (int i = 0; i <= nums.length; i++) {
curr = (i == nums.length) ? upper + 1 : nums[i];
if (prev + 2 <= curr) {
rst.add(printRange(prev + 1, curr - 1));
}
prev = curr;
}
return rst;
}
public String printRange(int from, int to) {
return (from == to) ? String.valueOf(from) : from + "->" + to;
}
}
```