Problem:
Given an ordered integer array nums with no repeated elements.
Returns a list of the smallest ordered range ranges that happen to cover all the numbers in the array. In other words, each element of nums is exactly covered by a certain range, and there is no number x belonging to a certain range but not belonging to nums.
Each interval range [a,b] in the list should be output in the following format:
"a->b", if a != b
"a", if a == b
Example 1:
输入：nums = [0,1,2,4,5,7]
输出：["0->2","4->5","7"]
解释：区间范围是：
[0,2] --> "0->2"
[4,5] --> "4->5"
[7,7] --> "7"
Example 2:
输入：nums = [0,2,3,4,6,8,9]
输出：["0","2->4","6","8->9"]
解释：区间范围是：
[0,0] --> "0"
[2,4] --> "2->4"
[6,6] --> "6"
[8,9] --> "8->9"
Example 3:
输入：nums = []
输出：[]
Example 4:
输入：nums = [-1]
输出：["-1"]
Example 5:
输入：nums = [0]
输出：["0"]
Idea:
Just traverse once
Code:
class Solution {
public:
vector<string> summaryRanges(vector<int>& nums) {
vector<string> ans;
int left = 0;
for (int i = 0; i < nums.size(); ++i) {
if (i + 1 == nums.size() || nums[i] + 1 != nums[i + 1]) {
ans.push_back(std::to_string(nums[left]) + (left == i ? "" : "->" + std::to_string(nums[i])));
left = i + 1;
}
}
return ans;
}
};