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. 神马复杂度为n^2,n^3的解法肯定是会超时的,ok目标为O(n); 以max[i]表示从任意...
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: 6Explanation: [2,3] has the largest product 6.Example 2: Input: [-2,0,-1]Output: 0Explanation: The result...
g(k) = Smallest product subarray, from index 0 up to k. Then, f(k) = max( f(k-1) * A[k], A[k], g(k-1) * A[k] ) g(k) = min( g(k-1) * A[k], A[k], f(k-1) * A[k] ) publicintmaxProduct(int[] nums) {if(nums ==null|| nums.length == 0) {return0...
Find the contiguous subarray within an array (containing at least one number) which has the largest product. [2,3,-2,4], the contiguous subarray[2,3]has the largest product =6. 思路: c[i]表示从0~i的数组中 包含nums[i]这个结点的最大乘积。状态转移方程:c[i] = max{nums[i],c[i-1...
Explanation: The result cannot be 2, because [-2,-1] is not a subarray. 1. 2. 3. 题目大意: 求出最大连续子数组乘积。 解题思路 —— 暴力求解 直接将所有的存在可能都求解出来,找出最大值。 AC代码 class Solution { public: int maxProduct(vector<int>& nums) { ...
g(k) = Smallest product subarray, from index 0 up to k. Then, f(k) = max( f(k-1) * A[k], A[k], g(k-1) * A[k] ) g(k) = min( g(k-1) * A[k], A[k], f(k-1) * A[k] ) There we have a dynamic programming formula. Using two arrays of size n, we could...
0149 Max Points on a Line 15.80% Hard 0150 Evaluate Reverse Polish Notation Go 32.50% Medium 0151 Reverse Words in a String 17.00% Medium 0152 Maximum Product Subarray Go 29.40% Medium 0153 Find Minimum in Rotated Sorted Array 43.10% Medium 0154 Find Minimum in Rotated Sorted Array ...
0149 Max Points on a Line 21.7% Hard 0150 Evaluate Reverse Polish Notation Go 44.1% Medium 0151 Reverse Words in a String Go 30.2% Medium 0152 Maximum Product Subarray Go 34.9% Medium 0153 Find Minimum in Rotated Sorted Array Go 48.4% Medium 0154 Find Minimum in Rotated Sorted Arra...
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.</pre> 思路: 一开始的暴力解法,找出所有的子数组,算出乘积,然后找出比较大的一个。但是复杂度是n2;换思路,dp来做,要用两个dp数组。其中f[i]表示[0,i]范围内并且包含nums[i]的最大子数组乘积。g[i]标识[0,i]范围内并且...
(i-1)%2# x 表示当前最大或最小下标 y 表示前面最大或最小下标imax[x]=max(imax[y]*nums[i],imin[y]*nums[i],nums[i])# nums[i]可能为负数,若为负数 前面最小 * nums[i]变为最大,前面最大 * nums[i]变为最小imin[x]=min(imax[y]*nums[i],imin[y]*nums[i],nums[i])res=max(...