标签：Plan result log style xpl sum from could rom
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.
Idea 1. How can we extend a solution from nums[0..i] to nums[0..i+1]? what shall we do for a newly added element nums[i+1]. Similar to Maximum Subarray LT53, assume we know the largest product for array nums[0..i],
if nums[i+1] >= 0, maxProduct(i+1) = Math.max(maxProduct(i) * nums[i+1], nums[i+1])
if nums[i+1] < 0, to get the max product ending at (i+1), we also need to know the min element ending at i which could be negative, maxProduct(i+1) = Math.max(minProduct(i) * nums[i+1], nums[i+1]).
For any element at i+1, maxProduct(i+1) = Math.max( nums[i+1], minProduct(i) * nums[i+1], maxProduct(i) * nums[i+1] )
Time complexity: O(n)
Space complexity: O(1)
1 class Solution { 2 public int maxProduct(int[] nums) { 3 int minHere = 1, maxHere = 1, result = Integer.MIN_VALUE; 4 5 for(int num: nums) { 6 int res1 = minHere * num; 7 int res2 = maxHere * num; 8 if(res1 < res2) { 9 minHere = Math.min(res1, num); 10 maxHere = Math.max(res2, num); 11 } 12 else { 13 minHere = Math.min(res2, num); 14 maxHere = Math.max(res1, num); 15 } 16 result = Math.max(result, maxHere); 17 } 18 return result; 19 } 20 }
1 class Solution { 2 public int maxProduct(int[] nums) { 3 int minHere = 1, maxHere = 1, result = Integer.MIN_VALUE; 4 5 for(int num: nums) { 6 int res1 = minHere * num; 7 int res2 = maxHere * num; 8 9 minHere = Math.min(Math.min(res1, res2), num); 10 maxHere = Math.max(Math.max(res1, res2), num); 11 12 result = Math.max(result, maxHere); 13 } 14 return result; 15 } 16 }
Maximum Product Subarray LT152
标签：Plan result log style xpl sum from could rom
原文地址：https://www.cnblogs.com/taste-it-own-it-love-it/p/10425208.html
