leetcode - Jump Game II

AkibaTakuya

于 2014-10-26 13:08:38 发布

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本文链接：https://blog.youkuaiyun.com/AkibaTakuya/article/details/40476111

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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位置到达最后一个位置所需的最少跳跃次数。通过动态规划的方法解决该问题，并提供了一个具体的示例进行说明。

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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.)

class Solution {
public:
    int jump(int A[], int n) {
        int max = 0;
		int *dp = new int[n];
        memset(dp, 0, sizeof(int) * n);
        for (int i = 0; i < n; i++) {
            for (int j = max - i + 1; j <= A[i] && i + j < n; j++) {
                dp[i + j] = dp[i] + 1;
            } 
            max = max > A[i] + i ? max : A[i] + i;
        }
        return dp[n - 1];
    }
};