leetcode 55. 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.

一开始用的双重for循环，但是提交的时候报了个超时的错误，

public class Solution {
    /**
     * @param A: A list of integers
     * @return: The boolean answer
     */
    public boolean canJump(int[] A) {
        boolean[] reachable = new boolean[A.length];
        reachable[0] = true;
        
        for (int i = 1; i< A.length; i++) {
            for (int j = 0; j < i; j++) {
                if (reachable[j] && j + A[j] >= i) {
                    reachable[i] = true;
                    break;
                }
            }
        }
        return reachable[A.length - 1];
    }
}

变成一个for循环提交成功。

i == max && A[i] == 0  的意思是当前这个就是目前走的对多的点了，但是他的值又是0，所以就没法再往下走了

／／／／／

实际上，如果能通过某种方法走到该数组的最后一个位置，那么这个数组的所有位置肯定都有办法走到

public boolean canJump(int[] A) {
    if(A.length <= 1)
        return true;
 
    int max = A[0]; //max stands for the largest index that can be reached.
 
    for(int i=0; i<A.length; i++){
        //if not enough to go to next
        if(i == max && A[i] == 0) 
            return false;
 
        //update max    
        if(i + A[i] > max){
            max = i + A[i];
        }
 
        //max is enough to reach the end
        if(max >= A.length-1) 
            return true;
    }
 
    return false;    
}