leetcode 45. Jump Game II

﹏ヽ暗。殇╰゛Y 2022-08-21 04:44 4阅读 0赞

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.

Your goal is to reach the last index in the minimum number of jumps.

For example:  
Given array A = `[2,3,1,1,4]`

The minimum number of jumps to reach the last index is `2`. (Jump `1` step from index 0 to 1, then `3` steps to the last index.)

Note:  
You can assume that you can always reach the last index.

超时

class Solution {
    	bool jump_once(vector<int>& nums, vector<int>&startpos, set<int>&history)
    	{
    		vector<int>newstartpos;
    		for (int i = 0; i < startpos.size(); i++)
    		{
    			if (nums[startpos[i]] >= nums.size() - 1 - startpos[i])
    				return true;
    			if (history.find(startpos[i] + nums[startpos[i]]) == history.end())
    			{
    				history.insert(startpos[i] + nums[startpos[i]]);
    				newstartpos.push_back(startpos[i] + nums[startpos[i]]);
    			}
    			int bb = startpos[i] ;
    			while (bb + 1 <= startpos[i] + nums[startpos[i]])
    			{
    				while (bb + 1 <= startpos[i] + nums[startpos[i]] && nums[startpos[i]] >= nums[bb + 1] + bb + 1 - startpos[i])
    					bb++;
    				if (bb + 1 != startpos[i] + nums[startpos[i]] + 1)
    					if (history.find(bb + 1) == history.end())
    					{
    						history.insert(bb + 1);
    						newstartpos.push_back(bb + 1);
    					}
    				bb++;
    			}
    		}
    		startpos = newstartpos;
    		return false;
    	}
    public:
    	int jump(vector<int>& nums) {
    		if (nums.size() == 1)
    			return 0;
    		bool f = false;
    		vector<int>startpos;
    		set<int>history;
    		startpos.push_back(0);
    		int k = 0;
    		while (!f)
    		{
    			f = jump_once(nums, startpos, history);
    			k++;
    		}
    		return k;
    	}
    };

accepted