[Algorithm] DP - Min Number of Jumps

标签：index Min -- jumps Number array Jumps DP minNumberOfJumps

You're given a non-empty array of positive integers where each integer represents the maximum number of steps you can take forward in the array. For example, if the element at index 1 is 3, you can go from index 1 to index 2, 3, or 4.

Write a function that returns the minimum number of jumps needed to reach the final index.

Note that jumping from index i to index i + x always constitutes one jump, no matter how large x is.

Sample Input

array = [3, 4, 2, 1, 2, 3, 7, 1, 1, 1, 3]

Sample Output

4 // 3 --> (4 or 2) --> (2 or 3) --> 7 --> 3

  For each item i, check all previous items, calcuate the min jumps needed to that item.

// T: O(N ^ 2)
function minNumberOfJumps(array) {
  const jumps = array.map(_ => Infinity)
  jumps[0] = 0;

  for (let i = 1; i < array.length; i++) {
    for (let j = 0; j < i; j++) {
      if (array[j] + j >= i) {
        jumps[i] = Math.min(jumps[i], jumps[j] + 1)
      }
    }
  }

  return jumps[jumps.length - 1]
}

// Do not edit the line below.
exports.minNumberOfJumps = minNumberOfJumps;

 

标签：index,Min,--,jumps,Number,array,Jumps,DP,minNumberOfJumps
From： https://www.cnblogs.com/Answer1215/p/16773507.html