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This article mainly explains "how to solve the jump game algorithm problem with JavaScript". Interested friends may wish to have a look. The method introduced in this paper is simple, fast and practical. Next let the editor to take you to learn "JavaScript how to solve the jump game algorithm problem" bar!

Topic: given an array of non-negative integers, you are initially in the first position of the array. Each element in the array represents the maximum length you can jump at that location. Judge whether you can reach the last position.

Example 1:

Input: [2, 3, 1, 1, 1, 4]

Output: true

Explanation: jump 1 step from position 0 to 1, then jump 3 steps to the last position.

Example 2:

Input: [3, 2, 1, 0, 4]

Output: false

Explanation: in any case, you will always reach the position with an index of 3. But the maximum jump length for this position is 0

So you'll never get to the last position.

Through observation, it is found that:

If there is no 0 in the array, you can definitely skip to the end. If there is a 0 in the array, the following conditions must be met in order to jump to the end, and the 0 must be skipped from a position before 0.

For example, the array in example 2 can skip to the end if it is:

1 、 [4,2,1,0,4]

2 、 [3,3,1,0,4]

3 、 [3,2,2,0,4]

What pattern did you find? The value of the player's position must be greater than the distance between the 0 location index and the current location index.

4 > 3-0, the index value of position 4 is 0, and the difference of index value of distance 0 is 3 > 3, so we can skip 0, and so on.

3 > 3-1

2 > 3-2

After finding the core solution to the problem, the general idea is to find out the position of all zeros in the array and determine whether all the numbers before that position can skip the zero position. The code is as follows:

Var canJump = function (nums) {

Var canJump0List = []

For (var I = 0; I

= 0;-- j) {

/ / the value of the player's position must be greater than the distance between the location index value of 0 and the current location index value.

If (nums [j] > I-j) {

CanJump0 = true

Break

}

}

/ / there may be multiple zeros in the array; so the results of all zero judgments are put into one array.

CanJump0List.push (canJump0)

}

}

/ / finally, the array is judged. As long as one does not meet the condition and returns false, it cannot skip to the last.

For (var I = 0; I < canJump0List.length; + + I) {

If (! canJump0List [I]) {

Return false

}

}

Return true

At this point, I believe you have a deeper understanding of "how to solve the jump game algorithm problem with JavaScript". You might as well do it in practice! Here is the website, more related content can enter the relevant channels to inquire, follow us, continue to learn!

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