(Java) LeetCode 152. Maximum Product Subarray —— 乘积最大子序列

Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product.

Example 1:

Input: [2,3,-2,4]
Output: 6
Explanation: [2,3] has the largest product 6.

Example 2:

Input: [-2,0,-1]
Output: 0
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.

暴力解法应该会超时吧，直接考虑用动态规划。用dp[i]来存储以nums[i]为结尾的数组的子序列的最大乘积。有一个问题是根据dp[i-1]求dp[i]时，需要判断nums[i]对全局的影响。因为符号对乘法的影响会使结果不能作为局部最优解，但仍然保留着成为全局最优解的可能。所以同时需要用两个变量存储当前最大的正数乘积，以及最小的负数乘积。当遍历到nums[i]时需要同时更新dp[i]，maxPositive以及minNegative。具体的递推是，更新maxPositive，minNegative，之后dp[i]取dp[i-1]和maxPositive中的最大值。更新最大正乘积以及最小负乘积时要记得考虑只有元素自己的情况，并且要分别独立更新。当然本题不需要存储dp[0] ~ dp[i-1]，因为dp[i]严格的只和dp[i-1]有关，所以只需要一个dp来存储上一个位置即可，空间可以优化成常数。


Java

class Solution {
    public int maxProduct(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        int maxP = nums[0], minN = nums[0], dp = nums[0];
        for (int i = 1; i < nums.length; i++) {
            int locMax = nums[i] * maxP, locMin = nums[i] * minN;
            maxP = Math.max(nums[i], Math.max(locMax, locMin));
            minN = Math.min(nums[i], Math.min(locMax, locMin));
            dp = Math.max(dp, maxP);
        }
        return dp;
    }
}

原文地址：https://www.cnblogs.com/tengdai/p/9281299.html

时间： 2024-12-28 02:26:14

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