55. Jump Game-优快云博客

本文链接：https://blog.youkuaiyun.com/daimingyang123/article/details/78307338

本文介绍了一个经典的算法问题——跳跃游戏。通过分析问题背景，提出了一种高效的贪心算法解决方案，并详细解释了其实现思路和步骤。该算法的目标是判断能否从数组的起始位置达到末尾。

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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.

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.

This is a dynamic programming[1] question. Usually, solving and fully understanding a dynamic programming problem is a 4 step process:

Start with the recursive backtracking solution
Optimize by using a memoization table (top-down[3] dynamic programming)
Remove the need for recursion (bottom-up dynamic programming)
Apply final tricks to reduce the time / memory complexity

All solutions presented below produce the correct result, but they differ in run time and memory requirements.

第一次就直接写出了3 还不错这里的贪心算法不是很难可以看看

public boolean canJump(int[] nums) {
        int lastPos = nums.length - 1;
        for (int i = nums.length - 1; i >= 0; i--) {
            if (i + nums[i] >= lastPos) {
                lastPos = i;
            }
        }
        return lastPos == 0;
    }

每次尝试最右的点跳最大步数看是否能达到 lastPos

一开始纠结会不会忽略情况比如

5 3 2 1 4

1可以到达4 2也可以 1会被作为lastPos 那后面就是找能到达1的点会不会忽略到达2的情况实际是不会的

2可以到达4 那就可以到达1 所以2就会被选作下一次的lastPos

相当于2->4被拆成了 2->1->4 大步被拆成小步最终效果是一样的