Problem45.Jump Game II

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#leetcode

本文介绍了一个算法问题，即如何在给定的非负整数数组中找到从第一个元素到最后一个元素所需的最小跳跃次数。该算法使用了动态规划的思想，通过维护当前能够达到的最大位置和已经完成的跳跃次数来解决问题。

问题：Given an array of non-negative integers, 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.

Example:

Input: [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.
设置count为走的步数，设置max,为当前位置能到达的最大位置，last,为step走的最长距离

class Solution {
    public int jump(int[] nums) {
        int n=nums.length;
        int count=0;
        int max=0,i=0,last=0;
        for(;i<n;i++){
            if(i>last){
                last=max;
                count++;
            }
            max=Math.max(max,i+nums[i]);
        }

        return count;
    }
}
//Runtime: 6 ms
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