LeetCode 152. Maximum Product Subarray

wenyq72026-10-29 9:22

Given an integer array nums, find asubarray 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.

Note that the product of an array with a single element is the value of that element.

Example 1:

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

Example 2:

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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 subarray of nums is guaranteed to fit in a 32-bit integer.

自己做就是不会，虽然其实挺直观的。

思路和53有点像，需要记录遍历到当前的最大值，以及最后的结果。但因为这题是乘法，可能会出现负负得正的情况，所以有可能负负得正以后反而成为了更大值，因此需要同时记录当前的最大和当前的最小。对于遍历到当前这个数字i，它这会儿的最大值就是max(nums $i$ , nums $i$ * currMax, nums $i$ * currMin)，以及需要更新currMax和currMin。

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class Solution {
    public int maxProduct(int[] nums) {
        int currMin = nums[0];
        int currMax = nums[0];
        int result = nums[0];

        for (int i = 1; i < nums.length; i++) {
            int newMin = Math.min(nums[i], Math.min(currMin * nums[i], currMax * nums[i]));
            int newMax = Math.max(nums[i], Math.max(currMin * nums[i], currMax * nums[i]));

            currMin = newMin;
            currMax = newMax;

            result = Math.max(result, currMax);
        }

        return result;        
    }
}

根据gpt老师，还可以简化成如果当前数字是负数，就交换最大值和最小值。我自己没写代码，太懒了，贴gpt老师的答案好了。

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class Solution {
    public int maxProduct(int[] nums) {
        int maxProduct = 1;
        int minProduct = 1;
        int result = nums[0];

        for (int x : nums) {
            if (x < 0) {
                int temp = maxProduct;
                maxProduct = minProduct;
                minProduct = temp;
            }

            maxProduct = Math.max(x, maxProduct * x);
            minProduct = Math.min(x, minProduct * x);

            result = Math.max(result, maxProduct);
        }

        return result;
    }
}