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.

一刷
Greedy貪婪法。維護(hù)一個(gè)maxCover，對(duì)數(shù)組從0 - maxCover或者nums.length進(jìn)行遍歷。當(dāng)maxCover >= nums.length - 1時(shí)表示可以跳到最后一個(gè)元素。需要計(jì)算精確。

Time Complexity - O(n)， Space Complexity - O(1)。

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

二刷
greedy, 維護(hù)一個(gè)maxCover

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