标签:ace HERE Plan ret pre out continue string const
You are given a sorted unique integer array nums
.
Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums
is covered by exactly one of the ranges, and there is no integer x
such that x
is in one of the ranges but not in nums
.
Each range [a,b]
in the list should be output as:
"a->b"
ifa != b
"a"
ifa == b
Example 1:
Input: nums = [0,1,2,4,5,7] Output: ["0->2","4->5","7"] Explanation: The ranges are: [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"] Explanation: The ranges are: [0,0] --> "0" [2,4] --> "2->4" [6,6] --> "6" [8,9] --> "8->9"
Example 3:
Input: nums = [] Output: []
Example 4:
Input: nums = [-1] Output: ["-1"]
Example 5:
Input: nums = [0] Output: ["0"]
Constraints:
0 <= nums.length <= 20
-231 <= nums[i] <= 231 - 1
- All the values of
nums
are unique.
two pointers, time = O(n), space = O(1)
class Solution { public List<String> summaryRanges(int[] nums) { List<String> res = new ArrayList<>(); if(nums == null || nums.length == 0) { return res; } for(int i = 0, j = 0; j < nums.length; j++) { if(j + 1 < nums.length && nums[j + 1] - nums[j] == 1) { continue; } if(i == j) { res.add(nums[i] + ""); } else { res.add(nums[i] + "->" + nums[j]); } i = j + 1; } return res; } }
228. Summary Ranges - Easy
标签:ace HERE Plan ret pre out continue string const
原文地址:https://www.cnblogs.com/fatttcat/p/13896476.html
暂无评论...