Leetcode刷题99-55. 跳跃游戏(C++详细解法！！！)

Come from : [https://leetcode-cn.com/problems/jump-game/]

1.Question

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.

Example 1:

Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
             jump length is 0, which makes it impossible to reach the last index.

2.Answer

meduim 类型题目。贪心算法。
算法分析：
这个题目的关键点是要 构建出 index数组（似不似对这个index数组有疑问，详见下图）。
具体步骤见下图：

AC代码如下：

class Solution {
public:
    bool canJump(vector<int>& nums) {
        vector<int> index;
        for(int i = 0; i < nums.size(); ++i)
        {
            index.push_back(nums[i] + i);  //计算index数组
        }

		//初始化
        int jump = 0;
        int max_index = index[0];
        
        //退出循环的条件：直到jump跳至数组尾部，或者jump超越了当前可以跳的最远位置
        while(jump < nums.size() && jump <= max_index )
        {
            if(max_index < index[jump])
            {
                max_index = index[jump];
            }
            ++jump;
        }

		//做出最后的判断
        if(jump == nums.size())
        {
            return true;
        }
        return false;
    }
};

3.大神的算法

大神AC速度第一代码：

class Solution {
public:
    bool canJump(vector<int>& nums) {
        int len = nums.size();
        if(len == 1)    return true;
        int max_len = 1;
        for(int i = 0; i < len - 1; i++)
        {
            if(i < max_len)
            {
                max_len = max(max_len, i + 1 + nums[i]);
            }
        }
        if(max_len >= len)  return true;
        return false;
    }
};

4.我的收获

贪心算法初步。。。
fighting。。。

2019/6/11 胡云层 于南京 99