228. Summary Ranges.md

228. Summary Ranges

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Difficulty : Easy

Related Topics : Array

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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
• -2^31 <= nums[i] <= 2^31 - 1
• All the values of nums are unique.

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Solution

• mine
  • Java
    • Runtime: 5 ms, faster than 86.58%, Memory Usage: 37.7 MB, less than 8.82% of Java online submissions

      public List<String> summaryRanges(int[] nums) {
          List<String> res = new ArrayList<>();
          int n = nums.length;
          if(n == 0) return res;
          if(n == 1){
              res.add(String.valueOf(nums[0]));
              return res;
          }
          int s = nums[0];
          int t = s;
          for(int i = 1; i < n; i++){
              if(i + 1 == n){
                  if(nums[i] == t + 1){
                      res.add(s + "->" + nums[i]);
                  }else{
                      if(s == t){
                          res.add(String.valueOf(s));
                      }else{
                          res.add(s + "->" + t);
                      }
                      res.add(String.valueOf(nums[i]));
                  }
              }else{
                  if(nums[i] == t + 1){
                      t++;
                  }else{
                      if(s == t){
                          res.add(String.valueOf(s));
                      }else{
                          res.add(s + "->" + t);
                      }
                      s = nums[i];
                      t = s;
                  }
              }
          }
          return res;
      }
      

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• the most votes

• Runtime: 0 ms, faster than 100.00%, Memory Usage: 37.2 MB, less than 8.82% of Java online submissions

  public List<String> summaryRanges(int[] nums) {
      List<String> res = new ArrayList<>();
      if (nums == null || nums.length == 0) {
          return res;
      }
      int anchor = 0;
      for (int i = 0; i < nums.length; i ++) {
          anchor = i;
          while(i + 1 < nums.length && nums[i + 1] - nums[i] == 1) {
              i ++;
          }
          res.add(getRange(nums[anchor], nums[i]));
      }
      return res;
  }
  
  private String getRange(int start, int end) {
      StringBuilder sb = new StringBuilder();
      if (start == end) {
          sb.append(start);
      } else {
          sb.append(start).append("->").append(end);
      }
      return sb.toString();
  }
  

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