LeetCode 44 Jump Game II

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本文探讨了如何解决在给定数组中找到最小跳跃次数以达到最后一个元素的问题。通过递归方法，从第一个元素开始，不断寻找可以跳到目标位置的最小步数。实例分析了如何将数组 [2,3,1,1,4] 跳跃到终点所需的最少步骤，并解释了算法背后的逻辑。

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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.)

思路：使用递归，始终去寻找最小的i能到目标位置。

class Counter {
	public static int counter;
}

public class Solution {
	public int jump(int[] A) {
		Counter.counter = 0;
		return jump(A, A.length - 1);
	}

	private int jump(int[] A, int end) {
		int i = 0;
		for (; i < end; i++) {
			if (i + A[i] >= end) {
				Counter.counter++;
				break;
			}
		}
		if (i == 0)
			return Counter.counter;
		jump(A, i);
		return Counter.counter;
	}
}