228. Summary Ranges - Easy

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" if a != b
  • "a" if a == 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

版权声明:完美者 发表于 2020-10-30 11:55:20。
转载请注明:228. Summary Ranges - Easy | 完美导航

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