Leetcode Jump game I, II

最新推荐文章于 2025-05-22 14:52:23 发布

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最新推荐文章于 2025-05-22 14:52:23 发布

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分类专栏： leetcode leetcode greedy

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本文解析了JumpGame I 和 II两个问题的算法实现，分别使用动态规划和贪心算法来判断是否能够到达数组最后一个位置及如何用最少步数到达终点。提供了详细的代码示例。

Jump game I

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.
Determine if you are able to reach the last index.

For example:
A = [2,3,1,1,4], return true.

A = [3,2,1,0,4], return false.

[code]

public class Solution {
    public boolean canJump(int[] A) {
        if(A==null || A.length<2)return true;
        boolean dp[]=new boolean [A.length];
        Arrays.fill(dp, false);
        dp[0]=true;
        for(int i=1;i<A.length;i++)
        {
            int j=i-1;
            while(j>=0)
            {
                if(dp[j]==true && A[j]>=i-j)
                {
                    dp[i]=true;break;
                }
                j--;
            }
            if(j<0)dp[i]=false;
        }
        return dp[dp.length-1];
    }

}

Jump game II

reach the last index in the minimum number of jumps.

[code]

public class Solution {
    public int jump(int[] A) {
        if(A==null || A.length<2)return 0;
        int start=0, end=0, jumps=0;
        while(start<=end && end<A.length-1)
        {
            int max=end;
            for(int i=start;i<=end;i++)
            {
                max=Math.max(max, i+A[i]);
            }
            start=end+1;
            end=max;
            jumps++;
        }
        return jumps;
    }

}