leetcode【228】【tag Array】Summary Ranges【c++版本，时间100%，有截图】

问题描述：

Given a sorted integer array without duplicates, return the summary of its ranges.

Example 1:

Input:  [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Explanation: 0,1,2 form a continuous range; 4,5 form a continuous range.

Example 2:

Input:  [0,2,3,4,6,8,9]
Output: ["0","2->4","6","8->9"]
Explanation: 2,3,4 form a continuous range; 8,9 form a continuous range.

源码：

我自己的下做法，很冗余，就是用Left和right存储左右边界。如果left = right-1说明只有一个元素。

时间100%，空间34%

class Solution {
public:
    vector<string> summaryRanges(vector<int>& nums) {
        int n = nums.size();
        vector<string> result;
        if(n==0)    return result;
        if(n==1){
            result.push_back(to_string(nums[0]));  return result;
        }
        int left = 0, right = 1;
        for (right=1; right<n; right++){
            // cout<<right<<endl;
            while (right<n && nums[right] == nums[right-1] + 1)   right++;
            if(left != right-1){
                result.push_back(to_string(nums[left]) + "->" + to_string(nums[right-1]));
                left = right;
            }
            else {
                result.push_back(to_string(nums[left]));
                left = right;
            }
            if (right == n-1)   result.push_back(to_string(nums[n-1]));
        }
        return result;
    }
};

翻了一下Discuss区大佬的做法，思路都差不多，有个很简洁的方法。时间同样100%：

class Solution {
public:
    vector<string> summaryRanges(vector<int>& nums) 
    {
        vector<string> result;
        int n = nums.size();
        for (auto left(0), right(0); right<n; right++){
            if(right==n-1 || nums[right+1]!=nums[right]+1){
                result.emplace_back(left==right? to_string(nums[right]): to_string(nums[left])+"->"+to_string(nums[right]));
                left = right+1;
            }
        }
        return result;
    }
};

其中emplace_back和push_back的差别如下：

push_bach():

首先需要调用构造函数构造一个临时对象，然后调用拷贝构造函数将这个临时对象放入容器中，然后释放临时变量。

emplace_back():

这个元素原地构造，不需要触发拷贝构造和转移构造。