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题目

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”].

解答

这一题很简答，主要是把数字转化为字符串，可以使用itoa，也可以使用to_string，但是在leetcode上使用itoa，貌似没有包含这个，就自己写了一个。

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

class Solution1 {public:    vector
   
     summaryRanges(vector
    
     & nums) {        vector
     
       res;        if (nums.size() == 0) return res;        int i = 0;        string end_s = "";        string res_temp;        while (i<=nums.size() - 1)        {            int start = i;            char p[20];            itoa(nums[start], p, 10);            res_temp = p;            int end = i;            while (i
      
       ";                res_temp = res_temp + end_s;            }            res.push_back(res_temp);        }        return res;    }private:    void itoa(int num, char *str, int radix)//这里不考虑其他进制的    {        char index[] = "0123456789ABCDEF";        unsigned unum;        int i = 0, j, k;        if (radix == 10 && num<0)//10进制负数        {            unum = (unsigned)-num;//unsigned是为了考虑 INT_MIN这种情况            str[i++] = '-';        }        else            unum = (unsigned)num;        do {            str[i++] = index[unum % (unsigned)radix];            unum /= radix;        } while (unum);        str[i] = '\0';        if (str[0] == '-') k = 1;        else k = 0;        char temp;        for (j = k; j<=(i - 1)/ 2; j++)//反转        {            temp = str[j];            str[j] = str[i - 1 + k - j];            str[i - 1 + k - j] = temp;        }    }};
      
     
    
   

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