Jump Game

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.

Solution: DP

这道题是动态规划的题目，所用到的方法跟是在Maximum Subarray中介绍的套路，用“局部最优和全局最优解法”。我们维护一个到目前为止能跳到的最远距离，以及从当前一步出发能跳到的最远距离。局部最优local=A[i]+i，而全局最优则是global=Math.max(global, local)。递推式出来了，代码就比较容易实现了。因为只需要一次遍历时间复杂度是O(n)，而空间上是O(1)。

public class Solution {
    public boolean canJump(int[] A) {
        if(A == null || A.length == 0)
            return false;
        
        int reach = 0;
        for(int i = 0; i <= reach && i < A.length; i++) {
            reach = Math.max(A[i] + i, reach);
        }
        if(reach < A.length - 1)
            return false;
        
        return true;
    }
}

Or

public class Solution {
    public boolean canJump(int[] A) {
        if(A == null || A.length == 0)
            return false;
        
        int reach = 0;
        for(int i = 0; i <= reach && i < A.length; i++) {
            if(A[i] + i > reach) {
                reach = A[i] + i;
            }
            if(reach >= A.length - 1) {
                return true;
            }
        }
        return false;
    }
}