Given a sorted integer array where the range of elements are in the inclusive range [lower, upper], 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"].
这道题开始做的比较混乱，看了大神的思路才得以理解。此题以nums[i]-1考虑结点，检查low和nums[i]-1的差值，如果为0，就是low+""，如果不为0，就是low+"->"+nums[i]-1;然后给low赋值成nums[i]+1.遍历完数组之后，检查low和upper的关系。此题一定要注意overflow问题，要定义成long型！！代码如下：
public class Solution {
public List<String> findMissingRanges(int[] nums, int lower, int upper) {
List<String> res= new ArrayList<>();
long low = lower;
long up =upper;
for(int n:nums){
long cur = (long)n-1;
if(low==cur) res.add(cur+"");
else if(low<cur) res.add(low+"->"+cur);
low = (long)n+1;
}
if(low<up) res.add(low+"->"+up);
else if(low==up) res.add(up+"");
return res;
}
}
原文：http://www.cnblogs.com/codeskiller/p/6361134.html