【LeetCode】Jump Game 解题报告

标签：jump game   greedy   贪心   

【题目】

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.

题意：给定一个数组，从第一个元素开始，元素的值表示能够往后跳的最大距离，问这样一个数组能否跳到最后一个元素。

思路：贪心。

这好像是第一次写贪心，代码不是很好，所以下面只讨论两个网上比较简洁的代码。

【逆向思路】

来源：https://oj.leetcode.com/discuss/11422/simplest-o-n-solution-with-constant-space

public class Solution {
    public boolean canJump(int[] A) {
        int last = A.length - 1;
        for (int i = A.length - 2; i >= 0; i--) {
            if (i + A[i] >= last) {
                last = i;
            }
        }
        return (last <= 0);
    }
}

【正向思路】

来源：https://oj.leetcode.com/discuss/15567/linear-and-simple-solution-in-c

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

【LeetCode】Jump Game 解题报告

标签：jump game   greedy   贪心   

原文地址：http://blog.csdn.net/ljiabin/article/details/41820071