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.

题意:

求最大乘积子数组

思路:

1. the sign(I mean,  if it is positive or negative) influence the product value

2. when iterating each item, it can be postive or negative

3. we use max[i] to stand for max product of ith item

4. we use min[i] to stand for min product of ith item

5. so function rule:

max[i]:    max(max[i -1]*nums[i], min[i-1]* nums[i], nums[i])

min[i]:     min(max[i -1]*nums[i], min[i-1]* nums[i], nums[i])

[leetcode]152. Maximum Product Subarray最大乘积子数组-LMLPHP

[leetcode]152. Maximum Product Subarray最大乘积子数组-LMLPHP

[leetcode]152. Maximum Product Subarray最大乘积子数组-LMLPHP

[leetcode]152. Maximum Product Subarray最大乘积子数组-LMLPHP

code

  public int maxProduct(int[] nums) {
// initialize
int[] max = new int[nums.length];
int[] min = new int[nums.length];
max[0] = nums[0];
min[0] = nums[0];
int result = nums[0];
// max[i]: max product of ith item
// min[i]: min product of ith item
for(int i = 1; i < nums.length; i++){
max[i] = Math.max(Math.max(max[i-1] * nums[i] , min[i-1]*nums[i]), nums[i]);
min[i] = Math.min(Math.min(max[i-1] * nums[i] , min[i-1]*nums[i]), nums[i]);
result = Math.max(result, max[i]);
}
return result;
}
04-25 13:45