228 Summary Range

//额外声明：我发现不必把做的每一道题都在博客里更新，今后将更新我认为比较有意思的题目

Descroption:

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"].

 给定一个排过序的无重复整形数组，返回其连续分组。这个题目标注难度为easy，但我着实耗费了一些时间，最终的代码也不够清晰，我除将把自己的代码列出来外，会再展示一个讨论区里面我认为逻辑清楚的代码作为比较，以供自己不时参考。

下面是我自己的代码：

class Solution {
public:
 struct group{
        group(): begin(0), end(0), tag(false) {}
        void reset() {
            begin = 0;
            end = 0;
            tag = false;
        }
        int begin;
        int end;
        bool tag;
    };
    vector<string> summaryRanges(vector<int>& nums) {
        vector<group> vec;
        vector<string> s;
        if(nums.empty()) return s;
        int cur = nums[0] - 1;
        group p;
        p.begin = nums[0];
        for(auto it: nums) {
            if(it == cur + 1) {
                if(it != p.begin) {
                    p.end = it;
                    p.tag = true;
                }
            } else {
                vec.push_back(p);
                p.reset();
                p.begin = it;
            }
            cur = it;
        }
        vec.push_back(p);
        for(int i = 0; i < vec.size(); i++) {
            char t[30];
            sprintf(t, "%d", vec[i].begin);
            string ss(t, t + strlen(t));
            if(vec[i].tag) {
                ss += "->";
                char u[30];
                sprintf(u, "%d", vec[i].end);
                ss += string(u, u+ strlen(u));
            }
            s.push_back(ss);
        }
        return s;
    }
};

更好的代码：

class Solution {
public:
    string buildString(int rangeStart, int rangeEnd) {
        string s;

        char buffer [33];

        sprintf(buffer,"%d",rangeStart);
        s.append(buffer);

        if (rangeStart != rangeEnd)
        {
            s.append("->");

            sprintf(buffer,"%d",rangeEnd);
            s.append(buffer);
        }

        return s;
    }

    vector<string> summaryRanges(vector<int>& nums) {
        vector<string> o;
        int rangeStart = 0;
        int rangeEnd = 0;

        if (nums.size() != 0)
        {
            for(int i = 0; i < nums.size(); i++)
            {
                if(i == nums.size()-1)
                {
                    rangeEnd = i;
                    string s = buildString(nums[rangeStart],nums[rangeEnd]);
                    o.push_back(s);
                }
                else
                    if((nums[i+1] == nums[i]+1))
                    {
                        rangeEnd++;
                    }
                    else
                    {
                        string s = buildString(nums[rangeStart],nums[rangeEnd]);
                        o.push_back(s);
                        rangeStart = i+1;
                        rangeEnd = i+1;
                    }
            }
        }

        return o;
    }
};