55. Jump Game

这几天在公司里有点咸呀，嘿嘿。闲来无事，随便做做题

这是道贪心题，确定个贪心策略就完事儿了：只要没有0，就一定能过去，甭管怎么走，肯定能过去；有0的话，就往前找，只要能找到一个点的值能一下子跨过0，也就能过去，找不到就过不去～

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.

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.

 

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