Viiisit [LeetCode] - 228. Summary Ranges

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Problem - Summary Ranges

You are given a sorted unique integer array nums.

A range [a,b] is the set of all integers from a to b (inclusive).

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.

翻譯蒟蒻

給定一個已排序且獨特的整數陣列 nums，要找出最小的一組排序範圍列表，這些範圍要包含陣列中的所有數字。

• 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"

---

Solution - JavaScript

/**
 * @param {number[]} nums
 * @return {string[]}
 */
var summaryRanges = function(nums) {
  let record = [];
  for (i = 0; i < nums.length; i++) {
    let temp = nums[i];
    while (nums[i] + 1 === nums[i+1]) {
      i++;
    }
    if (temp !== nums[i]) {
      record.push(temp + '->' + nums[i]);
    } else {
      record.push(temp.toString());
    }
  }
  return record;
};

Solution - Ruby

def summary_ranges(nums)
  record = []
  i = 0

  while i < nums.size
    start = nums[i]
    while i + 1 < nums.size && nums[i] + 1 == nums[i + 1]
      i += 1
    end

    if start != nums[i]
      record << "#{start}->#{nums[i]}"
    else
      record << "#{start}"
    end

    i += 1
  end

  record
end

Solution - PHP

function summaryRanges($nums) {
  $record = [];
  $i = 0;

  while ($i < count($nums)) {
    $start = $nums[$i];
    $end = $nums[$i];

    while ($i + 1 < count($nums) && $nums[$i + 1] === $nums[$i] + 1) {
      $end = $nums[$i + 1];
      $i++;
    }

    if ($start === $end) {
      $record[] = $start . '';
    } else {
      $record[] = $start . '->' . $end;
    }

    $i++;
  }

  return $record;
}

LeetCode 傳送門 - Summary Ranges