Viiisit [LeetCode] - 45. Jump Game II

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Problem - Jump Game II

You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].

Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:

0 <= j <= nums[i] and
i + j < n
Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].

翻譯蒟蒻

找出從數組的第一個元素開始，按照每個元素所允許的最大跳躍步數，最終到達數組最後一個元素所需的最小跳躍次數。

• Example 1:

  Input: nums = [2,3,1,1,4]
  Output: 2
  Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.

• Example 2:

  Input: nums = [2,3,0,1,4]
  Output: 2

---

Solution - JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var jump = function(nums) {
  let currentPosition = 0; 
  let maxPosition = 0; 
  let steps = 0;

  for (let i = 0; i < nums.length - 1; i++) {
    // 在當前跳躍範圍內，更新能夠跳躍到的最遠位置
    maxPosition = Math.max(maxPosition, i + nums[i]);

    // 如果到達當前跳躍的最遠位置，更新下一個跳躍的最遠位置
    if (i === currentPosition) {
      currentPosition = maxPosition;
      steps++;
    }
  }

  return steps;
};

Solution - Ruby

def jump(nums)
  current_position = 0
  max_position = 0
  steps = 0

  # 使用迴圈遍歷 nums 中的元素，只迭代到倒數第二個元素
  (nums.length - 1).times do |i|
    current_position = [current_position, i + nums[i]].max
    if i == max_position
      max_position = current_position
      steps += 1                      
    end
  end

  steps
end

LeetCode 傳送門 - Jump Game II