55. Jump Game

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.

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 Array Greedy 

链接：http://leetcode.com/problems/jump-game/

题解：

Greedy贪婪法。维护一个maxCover，对数组从0 - maxCover或者nums.length进行遍历。当maxCover >= nums.length - 1时表示可以跳到最后一个元素。需要计算精确。

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;
    }
}

Update:

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(nums[i] + i, maxCover);
            if(maxCover >= nums.length - 1)
                return true;
        }
        
        return false;
    }
}

　　

二刷：

Java:

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++) {
            if (i + nums[i] > maxCover) {
                maxCover = i + nums[i];
            }
            if (maxCover >= nums.length - 1) {
                return true;
            }
        }
        return false;
    }
}

三刷:

Java:

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++) {
            if (i + nums[i] >= nums.length - 1) return true;
            maxCover = Math.max(maxCover, i + nums[i]);
        }
        return false;
    }
}