【leetcode】228. Summary Ranges

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

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思路：
思路很简单，就是从头到尾遍历一遍元素，看看是否与前一个元素差1。需要注意的是int的溢出问题，凡是涉及到元素运算的地方，都要先将元素转换成long类型，防止溢出。

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代码实现：

class Solution {
public:
    string itos(int i){
        string str;
        stringstream stream;
        stream << i;
        stream >> str;
        return str;
    }
    
    vector<string> summaryRanges(vector<int>& nums) {
        if (nums.size() <= 0){
            return {};
        }
        
        vector<string> ret;
        int begin = 0;
        for (int i = 1; i < nums.size(); ++i){
            long pre_num = nums[i-1];
            long cur_num = nums[i];
            if (cur_num-pre_num == 1){
                continue ;
            }else{
                if (begin == i-1){
                    string t = itos(nums[begin]);
                    ret.push_back(t);
                }else{
                    string t = itos(nums[begin])+"->"+itos(nums[i-1]);
                    ret.push_back(t);
                }
                begin = i;
            }
        }
        
        if (begin == nums.size()-1){
            string t = itos(nums[begin]);
            ret.push_back(t);
        }else{
            string t = itos(nums[begin])+"->"+itos(nums[nums.size()-1]);
            ret.push_back(t);
        }
        
        return ret;
    }
    
};

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