152. Maximum Product Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest product.

For example, given the array [2,3,-2,4],

the contiguous subarray [2,3] has the largest product = 6.

Solution 1

public static int maxProduct(int[] nums) {if (nums.length == 0) {return 0;}int maxherepre = nums[0];int minherepre = nums[0];int maxsofar = nums[0];int maxhere, minhere;for (int i = 1; i < nums.length; i++) {maxhere = Math.max(Math.max(maxherepre * nums[i], minherepre * nums[i]), nums[i]);minhere = Math.min(Math.min(maxherepre * nums[i], minherepre * nums[i]), nums[i]);maxsofar = Math.max(maxhere, maxsofar);maxherepre = maxhere;minherepre = minhere;}return maxsofar;}

Solution 2 Same as above

public int maxProduct3(int[] nums) {        int result = nums[0];        int max = nums[0];        int min = nums[0];        for(int i = 1; i < nums.length; i++){            int temp = max;            max = Math.max(Math.max(max * nums[i], min * nums[i]), nums[i]);            min = Math.min(Math.min(temp * nums[i], min * nums[i]), nums[i]);            if(max > result){                result = max;            }        }        return result;    }

Solution 3

public static int maxProduct2(int[] nums) {int n = nums.length;int r = nums[0];for (int i = 1, imax = r, imin = r; i < n; i++) {if(nums[i] < 0){int temp = imax;imax = imin;imin = temp;}imax = Math.max(nums[i], imax * nums[i]);imin = Math.min(nums[i], imin * nums[i]);r = Math.max(r, imax);}return r;}