Content

Given an integer array nums, find a subarray that has the largest product, and return the product.

The test cases are generated so that the answer will fit in a 32-bit integer.

Example 1:

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

Example 2:

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

Constraints:

1 <= nums.length <= 2 * 10⁴
-10 <= nums[i] <= 10
The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

Solution

1. 动态规划

Java

class Solution {
    public int maxProduct(int[] nums) {
        // 1 <= nums.length <= 2 * 10⁴
        // -10 <= nums[i] <= 10
        // The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit
        //integer.
        int n = nums.length;
        // dp[i][0]表示以下标为i的数字结尾的子数组的最大乘积
        // dp[i][1]表示以下标为i的数字结尾的子数组的最小乘积
        int[][] dp = new int[n][2];
        dp[0][0] = nums[0];
        dp[0][1] = nums[0];
        int max = nums[0];
        for (int i = 1; i < n; i++) {
            if (nums[i] == 0) {
                dp[i][0] = 0;
                dp[i][1] = 0;
            } else if (nums[i] > 0) {
                dp[i][0] = dp[i - 1][0] > 0 ? dp[i - 1][0] * nums[i] : nums[i];
                dp[i][1] = dp[i - 1][1] > 0 ? nums[i] : dp[i - 1][1] * nums[i];
            } else {
                dp[i][0] = dp[i - 1][1] > 0 ? nums[i] : dp[i - 1][1] * nums[i];
                dp[i][1] = dp[i - 1][0] > 0 ? dp[i - 1][0] * nums[i] : nums[i];
            }
            max = Math.max(max, dp[i][0]);
        }
        return max;
    }
}