【區間 dp】A012_LC_兩個子序列的最大點積（分類討論）

一、Problem

Given two arrays nums1 and nums2.

Return the maximum dot product between non-empty subsequences of nums1 and nums2 with the same length.

A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, [2,3,5] is a subsequence of [1,2,3,4,5] while [1,5,3] is not).

Input: nums1 = [2,1,-2,5], nums2 = [3,0,-6]
Output: 18
Explanation: Take subsequence [2,-2] from nums1 and subsequence [3,-6] from nums2.
Their dot product is (2*3 + (-2)*(-6)) = 18.

二、Solution

方法一：dp

兩個序列取別名爲 A、B

• 定義狀態：
  • $dp[i][j]$ 表示取 A 中的前 $i$ 個數字和 B 中的前 $j$ 個數字可以組成的最大點積和。
• 思考初始化：
  • $dp[i][j] = -INF$，因爲系列存在負數，所以結果可能爲負數。
• 思考狀態轉移方程：
  • dp[i][j] = A[i] × B[j]，只選擇 A、B 當前的兩個數，不選其它數
  • dp[i][j] = max(dp[i][j], dp[i-1][j])，只選 B 的第 $j$ 個數
  • dp[i][j] = max(dp[i][j], dp[i][j-1])，只選 A 的第 $i$ 個數
  • dp[i][j] = max(dp[i][j], dp[i-1][j-1] + A[i] × B[j])，選擇 A、B 當前的兩個數和前面的選的數組成的最長點積和。
• 思考輸出：$dp[n1][n2]$

class Solution {
    public int maxDotProduct(int[] A, int[] B) {
        int n1 = A.length, n2 = B.length;
        int max = 0, INF = -0x3f3f3f3f, dp[][] = new int[n1+1][n2+1];
        for (int i = 0; i <= n1; i++)
        for (int j = 0; j <= n2; j++)
            dp[i][j] = INF;
        
        for (int i = 1; i <= n1; i++)
        for (int j = 1; j <= n2; j++) {
            int pro =  A[i-1]*B[j-1];
            dp[i][j] = pro;
            dp[i][j] = Math.max(dp[i][j], dp[i-1][j]);
            dp[i][j] = Math.max(dp[i][j], dp[i][j-1]);
            dp[i][j] = Math.max(dp[i][j], dp[i-1][j-1] + pro);
        }
        return dp[n1][n2];
    }
}

複雜度分析

• 時間複雜度：$O(n1 × n2)$，
• 空間複雜度：$O(n1 × n2)$