LeetCode 1269. 停在原地的方案数 (Java 记忆化，动态规划)

Posted 2022-11-28 oyzg

Java 记忆化，动态规划

回溯+记忆化

class Solution 
    private final int m = 1000000007;
	//map用来记录
	private Map<String, Integer> map = new HashMap<String, Integer>();
	
	public int numWays(int steps, int arrLen) 
		return dfs(steps,arrLen,0);
    
	/**
	 * 
	 * @param steps 剩余步数
	 * @param arrLen 数组长度
	 * @param index 当前下标
	 * @return
	 */
	public int dfs(int steps, int arrLen, int index) 
		if(steps == 0) 
			return index == 0 ? 1 : 0;
		
		//如果下标比步数大，那么就不能回到原点了，直接返回0
		if(steps < index) return 0;
		String s = ""+steps+","+index;
		//如果之前已经有记录了，直接返回
		if(map.containsKey(s)) return map.get(s);
		//原地不动的情况
		int count = dfs(steps-1,arrLen,index);
		count %= m;
		//向左跳一步的情况，记得取模
		if(index > 0) 
			count += dfs(steps-1, arrLen, index-1);
			count %= m;
		
		//向右跳一步
		if(index < arrLen-1)  
			count += dfs(steps-1, arrLen, index+1);
			count %= m;
		
		//map记录
		map.put(s, count);
		return count;
	

动态规划

class Solution 
    public int numWays(int steps, int arrLen) 
        int m = 1000000007;
		int len = Math.min(arrLen, steps);
		//dp[i][j]表示剩i步时下标为j的方案数
		int[][] dp = new int[steps+1][len+1];
		//初始化
		dp[0][0] = 1;
		for(int i = 1; i <= steps; i++) 
			for(int j = 0; j < len && j <= i; j++) 
				dp[i][j] = dp[i-1][j];
				if(j > 0) 
					dp[i][j] += dp[i-1][j-1];
					dp[i][j] %= m;
				
				if(j < arrLen-1) 
					dp[i][j] += dp[i-1][j+1];
					dp[i][j] %= m;
				
				
			
		
		return dp[steps][0];