[leetcode] 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.

https://oj.leetcode.com/problems/jump-game/

思路1:从左向右扫描数组,在当前可达的范围内前进,并且不断更新max。

思路2:看到有人用dp做的,需要O(n)空间。

思路1代码:

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

            if (i <= canReach) { // This makes sure this i is currently reachable
                canReach = canReach > (i + A[i]) ? canReach : (i + A[i]);
            }

        }
        return false;

    }

    public static void main(String[] args) {
        System.out.println(new Solution().canJump(new int[] { 2, 3, 1, 1, 4 }));
        System.out.println(new Solution().canJump(new int[] { 3, 2, 1, 0, 4 }));
    }
}
原文地址:https://www.cnblogs.com/jdflyfly/p/3810767.html