Summary Ranges 有序数组找出连续的子数组并且输出

出处：http://blog.youkuaiyun.com/xudli/article/details/46645087

Given a sorted integer array without duplicates, return the summary of its ranges.

For example, given [0,1,2,4,5,7], return ["0->2","4->5","7"].

public class Solution {
    public List<String> summaryRanges(int[] nums) {
        List<String> ret=new ArrayList<String>();
        int start=0;
        int end=0;
        if(nums.length==0)  return ret;
        while(end<nums.length){
               if(end+1<nums.length && nums[end+1]==nums[end]+1) end++;//end+1<nums.length不能少，否则会产生越界
               else{
                   if(start==end)   ret.add(Integer.toString(nums[end]));
                   else{
                       String s=nums[start]+"->"+nums[end];
                       ret.add(s);
                   }
                   end++;
                   start=end;
               }
        }
        return ret;
 
    }
}