228. Summary Ranges

228. Summary Ranges

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Total Accepted: 36554  Total Submissions: 159532  Difficulty: Easy

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"].

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

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分析：

字符串朴素的模拟思想，思路很直白

class Solution {
public:
    vector<string> summaryRanges(vector<int>& nums) {
        vector<string> result;
        int i=0;
        while(i<nums.size())
        {
            string ans=to_string(nums[i]);
            bool flag=false;
            //一，添加箭头
            if( ++i < nums.size() && nums[i]==(nums[i-1]+1) )//箭头的依据
              {  
                  ans.append("->");
                  flag=true;//箭头已经添加
              }
            //二，寻找箭头后面那个数字
            while(flag && i<nums.size())
            {
                if(nums[i]!=nums[i-1]+1)
                   break;
                i++; 
            }
            //三，追加箭头后面的数
            if(flag)
                ans.append(to_string(nums[i-1]));
            result.push_back(ans);    
        }
        return result;
    }
};

学习别人的算法设计：

（双指针）

class Solution {
public:
    vector<string> summaryRanges(vector<int>& nums) {
        vector<string> result;
        for (int i = 0; i < nums.size();) //遍历数组
        {
            int start = i, end = i;//起始与终止变量
            while (end + 1 < nums.size() && nums[end+1] == nums[end] + 1) //寻找终止变量的位置
                end++;
            if (end > start) 
                result.push_back(to_string(nums[start]) + "->" + to_string(nums[end]));
            else 
                result.push_back(to_string(nums[start]));
            i = end+1;
        }
        return result;
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50649914

原作者博客：http://blog.csdn.net/ebowtang