leetcode-【中等题】228. Summary Ranges

题目：

228. Summary Ranges

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"].

答案：

就是找连续的序列。

直接判断相邻数据大小是否相差为1即可，主要是注意最后的数字要特殊考虑一下。

链接：

https://leetcode.com/problems/summary-ranges/

代码：

#include <vector>
#include <string>
#include <sstream>

using std::vector;
using std::string;
using std::ostringstream;

class Solution {
private:
    string num2str(int& num)
    {
        ostringstream stream;
        stream << num;
        return stream.str();
    }
    
public:
    vector<string> summaryRanges(vector<int>& nums) {
        unsigned int len = nums.size();
        unsigned int index,inIndex;
        unsigned int minIndex = 0;
        unsigned int maxIndex = 0;
        
        vector<string> ans;
        if(len == 0)
        {
            return ans;
        }
        
        if(len == 1)
        {
            ans.push_back(num2str(nums[0]));
            return ans;
        }
        
        for(index = 0; index < len - 1; ++ index)
        {
            if(nums[index] + 1 == nums[index + 1])
            {
                maxIndex = index + 1;
            }else
            {
                string tmp = "";
                tmp += num2str(nums[minIndex]);
                if(maxIndex > minIndex)
                {
                    tmp += "->";
                    tmp += num2str(nums[maxIndex]);
                }
                ans.push_back(tmp);
                
                minIndex = index + 1;
                maxIndex = index + 1;
            }
        }
        
        if(nums[len-2] + 1 == nums[len - 1])
        {
            maxIndex = len - 1;
        }else
        {
            minIndex = len - 1;
            maxIndex = len - 1;
        }
        
        string tmp = "";
        tmp += num2str(nums[minIndex]);
        if(maxIndex > minIndex)
        {
            tmp += "->";
            tmp += num2str(nums[maxIndex]);
        }
        ans.push_back(tmp);
                
        return ans;
    }
};