1458. Max Dot Product of Two Subsequences

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).

Example 1:

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.

Example 2:

Input: nums1 = [3,-2], nums2 = [2,-6,7]
Output: 21
Explanation: Take subsequence [3] from nums1 and subsequence [7] from nums2.
Their dot product is (3*7) = 21.

Example 3:

Input: nums1 = [-1,-1], nums2 = [1,1]
Output: -1
Explanation: Take subsequence [-1] from nums1 and subsequence [1] from nums2.
Their dot product is -1.

Constraints:

• 1 <= nums1.length, nums2.length <= 500
• -1000 <= nums1[i], nums2[i] <= 1000

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给定两个数组，问从这两个数组中选出两个长度相同的子序列，使得子序列的向量点积最大，求这个最大值。

难题，参考讨论区解法： https://leetcode.com/problems/max-dot-product-of-two-subsequences/discuss/648261/C++Python-Simple-DP 。

假设dp[i][j]表示nums1[0:i]和nums2[0:j]的最大点积。则在求解dp[i][j]时，有以下四种情况：

• 不选nums1第i个数，dp[i][j]=dp[i-1][j]
• 不选nums2第j个数，dp[i][j]=dp[i][j-1]
• 既不选nums1第i个数，也不选nums2第j个数，dp[i][j]=dp[i-1][j-1]
• 既选nums1第i个数，也选nums2第j个数，dp[i][j]=dp[i-1][j-1]+nums1[i]*nums2[j]

在最后一个情况中，如果dp[i-1][j-1]<0的话，就把前面的点积都扔掉，相当于max(dp[i-1][j-1],0)，完整代码如下：

class Solution {
public:
	int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {
		int m = nums1.size(), n = nums2.size();
		vector<vector<int>> dp(m + 1, vector<int>(n + 1, INT_MIN));
		for (int i = 1; i <= m; ++i) {
			for (int j = 1; j <= n; ++j) {
				dp[i][j] = max(dp[i][j], dp[i - 1][j]);
				dp[i][j] = max(dp[i][j], dp[i][j - 1]);
				dp[i][j] = max(dp[i][j], dp[i - 1][j - 1]);
				dp[i][j] = max(dp[i][j], max(dp[i - 1][j - 1], 0) + nums1[i - 1] * nums2[j - 1]);
			}
		}
		return dp[m][n];
	}
};

本代码提交AC，用时96MS。