Class Solution
Medium
Given an array of non-negative integers nums, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Your goal is to reach the last index in the minimum number of jumps.
You can assume that you can always reach the last index.
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
Constraints:
1 <= nums.length <= 1040 <= nums[i] <= 1000
To solve the “Jump Game II” problem in Java with a Solution class, we can follow these steps:
- Define a
Solutionclass. - Define a method named
jumpthat takes an array of non-negative integersnumsas input and returns the minimum number of jumps required to reach the last index. - Initialize variables
maxReach,steps, andendto keep track of the maximum reachable position, the number of steps taken, and the end position respectively. InitializemaxReachto 0 andendto 0. - Iterate through the array from index 0 to
nums.length - 2:- Update
maxReachas the maximum ofmaxReachandi + nums[i]. - If the current index
iequalsend, updateendtomaxReachand incrementsteps.
- Update
- Return
steps.
Here’s the implementation:
public class Solution {
public int jump(int[] nums) {
int maxReach = 0;
int steps = 0;
int end = 0;
for (int i = 0; i < nums.length - 1; i++) {
maxReach = Math.max(maxReach, i + nums[i]);
if (i == end) {
end = maxReach;
steps++;
}
}
return steps;
}
}
This implementation provides a solution to the “Jump Game II” problem in Java. It calculates the minimum number of jumps required to reach the last index by iterating through the array and updating the maximum reachable position and the end position accordingly.
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Constructor Summary
Constructors -
Method Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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jump
public int jump(int[] nums)
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