LeetCode刷题：403. Frog Jump 青蛙过河问题JAVA算法设计

LeetCode刷题：403. Frog Jump 青蛙过河问题JAVA算法设计

A frog is crossing a river. The river is divided into x units and at each unit there may or may not exist a stone. The frog can jump on a stone, but it must not jump into the water.

Given a list of stones' positions (in units) in sorted ascending order, determine if the frog is able to cross the river by landing on the last stone. Initially, the frog is on the first stone and assume the first jump must be 1 unit.

If the frog's last jump was k units, then its next jump must be either k - 1, k, or k + 1 units. Note that the frog can only jump in the forward direction.

Note:

The number of stones is ≥ 2 and is < 1,100.
Each stone's position will be a non-negative integer < 231.
The first stone's position is always 0.
Example 1:

[0,1,3,5,6,8,12,17]

There are a total of 8 stones.
The first stone at the 0th unit, second stone at the 1st unit,
third stone at the 3rd unit, and so on...
The last stone at the 17th unit.

Return true. The frog can jump to the last stone by jumping 
1 unit to the 2nd stone, then 2 units to the 3rd stone, then 
2 units to the 4th stone, then 3 units to the 6th stone, 
4 units to the 7th stone, and 5 units to the 8th stone.
Example 2:

[0,1,2,3,4,8,9,11]

Return false. There is no way to jump to the last stone as 
the gap between the 5th and 6th stone is too large.

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算法设计

动态规划算法

package com.bean.algorithm.dp;

public class FrogJumpDriver {

	public boolean canCross(int[] stones) {
		// dp[i][k] means we can jump k places from the ith stone. So ,basically,
		// we can reach the ith stone from any jth stone(j<i) when dp[j][distance
		// between i & j] is true.
		// and WHEN that happens,for that particular j&i pair, we can say that
		// dp[i][distance between i & j],
		// dp[i][distance between i & j + 1], dp[i][distance between i & j - 1] are
		// true.
		// so the 2d array will essentially hold the information of all possible
		// combinations of i & j (j<i)
		// along with the respective jumps it can make from there

		int n = stones.length;
		int dp[][] = new int[n + 1][n + 1]; // 0 is false, 1 is true

		// also,notice that since the starting k is 1, and due to the constraint
		// k,k-1,k+1, our dist variable
		// can never be greater that n+1

		dp[0][1] = 1; // givern condition

		for (int i = 1; i < n; i++) {
			for (int j = 0; j < i; j++) { // we are computing possibilities for every (j,i) pair

				int dist = stones[i] - stones[j];
				if (dist > 0 && dist < n + 1 && dp[j][dist] == 1) {
					dp[i][dist] = 1;
					dp[i][dist - 1] = 1;
					dp[i][dist + 1] = 1;
				}
			}
		}

		for (int k = 0; k < n; k++) {
			if (dp[n - 1][k] == 1)
				return true; // so for any dist k, if we can reach n-1 th stone, we return true
		}
		return false;
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FrogJumpDriver frogJump=new FrogJumpDriver();
		
		int[] stones=new int[] {0,1,3,5,6,8,12,17};
		boolean flag=frogJump.canCross(stones);
		System.out.println("Flag is: "+flag);
	}

}

程序运行结果：

Flag is: true

LeetCode提交，Accepted~