228. Summary Ranges

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

Example 1:

Input: [0,1,2,4,5,7]
Output: [“0->2”,“4->5”,“7”]

Example 2:

Input: [0,2,3,4,6,8,9]
Output: [“0”,“2->4”,“6”,“8->9”]
#思路
本题按照题意需要将数组中连续数字转换成用箭头表示的范围。观察Example1及Example2就能明白。思路比较简单。这里介绍一个O(n)时间复杂度的算法。维护两个变量，分别代表范围数字起始数字索引start及结尾数字索引end。然后扫描一遍数组，并按照"start->end”or "start"的形式保存在字符串数组中。
#代码

class Solution {
public:
    vector<string> summaryRanges(vector<int>& nums) {
        vector<string> res;
        if(nums.size()<=0)
            return res;
        int start = 0,end=0;
        for(int i=1;i<nums.size();i++)
        {
            if(nums[i-1]+1==nums[i])
            {
                end = i;
                continue;
            }
            else if(end>start)
                res.push_back(to_string(nums[start])+"->"+to_string(nums[end]));
            else
                res.push_back(to_string(nums[start]));    
            start = i;
            end = i;
        }
        if(end>start)
            res.push_back(to_string(nums[start])+"->"+to_string(nums[end]));
        else
            res.push_back(to_string(nums[start])); 
        return res;
    }
};