228 Summary Ranges 汇总区间
Description:
Given a sorted integer array without duplicates, return the summary of its ranges.
Example:
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.
题目描述:
给定一个无重复元素的有序整数数组，返回数组区间范围的汇总。
示例 :
示例 1:
输入: [0,1,2,4,5,7]
输出: ["0->2","4->5","7"]
解释: 0,1,2 可组成一个连续的区间; 4,5 可组成一个连续的区间。
示例 2:
输入: [0,2,3,4,6,8,9]
输出: ["0","2->4","6","8->9"]
解释: 2,3,4 可组成一个连续的区间; 8,9 可组成一个连续的区间。
思路:
双指针
右指针负责找到第一个 nums[r] + 1 != nums[r + 1]的位置
左指针指向上一个区间的右边 + 1的位置
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
vector summaryRanges(vector& nums)
{
vector result;
int l = 0, r = 0;
while (r < nums.size())
{
while (r < nums.size() - 1 and nums[r] + 1 == nums[r + 1]) ++r;
if (l != r) result.push_back(to_string(nums[l]) + "->" + to_string(nums[r]));
else result.push_back(to_string(nums[l]));
++r;
l = r;
}
return result;
}
};
Java:
class Solution {
public List summaryRanges(int[] nums) {
List result = new ArrayList<>();
int l = 0, r = 0;
while (r < nums.length) {
while (r < nums.length - 1 && nums[r] + 1 == nums[r + 1]) ++r;
if (l != r) result.add(String.valueOf(nums[l]) + "->" + String.valueOf(nums[r]));
else result.add(String.valueOf(nums[l]));
++r;
l = r;
}
return result;
}
}
Python:
class Solution:
def summaryRanges(self, nums: List[int]) -> List[str]:
if not nums:
return []
result = [[nums[0]]]
for i in range(1, len(nums)):
if nums[i] - nums[i - 1] != 1:
result[-1].append(nums[i - 1])
result.append([nums[i]])
result[-1].append(nums[-1])
return [[f'{a}->{b}', f'{a}'][a == b] for a, b in result]