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：

 1 class Solution {
 2 public:
 3  struct group{
 4         group(): begin(0), end(0), tag(false) {}
 5         void reset() {
 6             begin = 0;
 7             end = 0;
 8             tag = false;
 9         }
10         int begin;
11         int end;
12         bool tag;
13     };
14     vector<string> summaryRanges(vector<int>& nums) {
15         vector<group> vec;
16         vector<string> s;
17         if(nums.empty()) return s;
18         int cur = nums[0] - 1;
19         group p;
20         p.begin = nums[0];
21         for(auto it: nums) {
22             if(it == cur + 1) {
23                 if(it != p.begin) {
24                     p.end = it;
25                     p.tag = true;
26                 }
27             } else {
28                 vec.push_back(p);
29                 p.reset();
30                 p.begin = it;
31             }
32             cur = it;
33         }
34         vec.push_back(p);
35         for(int i = 0; i < vec.size(); i++) {
36             char t[30];
37             sprintf(t, "%d", vec[i].begin);
38             string ss(t, t + strlen(t));
39             if(vec[i].tag) {
40                 ss += "->";
41                 char u[30];
42                 sprintf(u, "%d", vec[i].end);
43                 ss += string(u, u+ strlen(u));
44             }
45             s.push_back(ss);
46         }
47         return s;
48     }
49 };

My Code

更好的代码：

 1 class Solution {
 2 public:
 3     string buildString(int rangeStart, int rangeEnd) {
 4         string s;
 5 
 6         char buffer [33];
 7 
 8         sprintf(buffer,"%d",rangeStart);
 9         s.append(buffer);
10 
11         if (rangeStart != rangeEnd)
12         {
13             s.append("->");
14 
15             sprintf(buffer,"%d",rangeEnd);
16             s.append(buffer);
17         }
18 
19         return s;
20     }
21 
22     vector<string> summaryRanges(vector<int>& nums) {
23         vector<string> o;
24         int rangeStart = 0;
25         int rangeEnd = 0;
26 
27         if (nums.size() != 0)
28         {
29             for(int i = 0; i < nums.size(); i++)
30             {
31                 if(i == nums.size()-1)
32                 {
33                     rangeEnd = i;
34                     string s = buildString(nums[rangeStart],nums[rangeEnd]);
35                     o.push_back(s);
36                 }
37                 else
38                     if((nums[i+1] == nums[i]+1))
39                     {
40                         rangeEnd++;
41                     }
42                     else
43                     {
44                         string s = buildString(nums[rangeStart],nums[rangeEnd]);
45                         o.push_back(s);
46                         rangeStart = i+1;
47                         rangeEnd = i+1;
48                     }
49             }
50         }
51 
52         return o;
53     }
54 };

Better Code