经典算法——Jump Game（II）

Posted 2020-10-01 cxchanpin

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.)

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;

class Solution {
public:
	int jump(vector<int> A)
	{
		int maxReach = A[0];
		int edge = 0; //edge表示当前能够达到最远的坐标边界值
		int minstep = 0;

		for (int i = 1; i < A.size(); i++)
		{
			//若当前坐标超过了最远的坐标边界值，应该跳跃一次，同一时候更新maxReach
			if (i > edge)
			{  
				minstep += 1;
				edge = maxReach;
				if (edge >= A.size() - 1) //若数组最后一个元素的坐标在edge覆盖的范围，则返回跳跃次数
					return minstep;
			}
			maxReach = max(maxReach, A[i] + i);
		}
		//假设不能达到数组最后一个元素，则返回0
		return 0;
	}
};


int main()
{
	Solution sol;
	vector<int> nums = {  5,9,3,2,1,0,2,3,3,1,0,0 };
	int res =sol.jump(nums);
	cout << res << endl;
	system("pause");
	return 0;
}