Title Description

  Given a No repeating elements Of Orderly An array of integers nums .
return Just over all the numbers in the array Of Minimum order List of ranges . in other words ,nums Every element of is just covered by an interval , And there's no existence that belongs to a certain range but doesn't belong to nums The number of x .
Each range in the list [a,b] It should be output in the following format ：

    1. "a->b" , If  a != b
    2. "a" , If  a == b

Example 1：

 Input ：nums = [0,1,2,4,5,7]
 Output ：["0->2","4->5","7"]
 explain ： The range is ：
[0,2] --> "0->2"
[4,5] --> "4->5"
[7,7] --> "7"

Example 2：

 Input ：nums = [0,2,3,4,6,8,9]
 Output ：["0","2->4","6","8->9"]
 explain ： The range is ：
[0,0] --> "0"
[2,4] --> "2->4"
[6,6] --> "6"
[8,9] --> "8->9"

Using double pointers to achieve , What needs to be noted is the condition of crossing the boundary .

vector<string> summaryRanges(vector<int>& nums) {
    
	vector<string> ans;
	string s;
	int len=nums.size();
	for(int i=0,j=0;j<len;j++){
    // Double pointer   We need to consider the problem of cross-border 
		if(j+1<len && nums[j]!=nums[j+1]-1){
    
			if(j-i==0){
    
				s=to_string(nums[i]);	
			}
			else{
    
				s=to_string(nums[i])+"->"+to_string(nums[j]);	
			}
			ans.push_back(s);
			s="";
			i=j+1;
		}
		else if(j+1==len){
    // The border has been reached 
			if(j-i==0){
    
				s=to_string(nums[i]);	
			}
			else{
    
				s=to_string(nums[i])+"->"+to_string(nums[j]);	
			}
			ans.push_back(s);	
		}
	}
	return ans;
}