LeetCode--------45. Jump Game II(跳跃游戏--最短路径)

最新推荐文章于 2022-07-26 21:49:37 发布

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本文介绍了一个算法问题：如何在给定的非负整数数组中，从第一个元素出发，每一步可以跳过最多该位置数值所指示的距离，最终达到数组最后一个元素所需的最少跳跃次数。文章详细解释了算法思路，并提供了一段Java代码实现。

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.

For example:
Given array A = [2,3,1,1,4]

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.)

Note:
You can assume that you can always reach the last index.

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jump:目前为止的jump数

maxS:从A[0]进行ret次jump之后达到的最大范围

maxC:从0~i这i+1个A元素中能达到的最大范围

当maxS < i，说明jump次jump已经不足以覆盖当前第i个元素，因此需要增加一次jump，使之达到

记录的maxC。

public class Solution {
    public int jump(int[] nums) {
        int jump=0;
        int maxS=0;
        int maxC=0;
        for(int i=0;i<nums.length;i++){
            if(maxS<i){
                jump++;
                maxS=maxC;
            }
            maxC=Math.max(maxC,nums[i]+i);
        }
        
       return jump; 
    }
}