intmaxProduct = A[0];//最大值的初始值 intmaxTemp = A[0];//累积的最大值 intminTemp = A[0];//累积的最小值 for(inti =1;i < A.length;i++) {//从数组的第二个元素开始遍历 inta = A[i]*maxTemp; intb = A[i]*minTemp; maxTemp = Math.max(Math.max(a,b), A[i]);//选取...
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...
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...
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...
LeetCode Maximum Product Subarray 最大子序列积 题意:给一个size大于0的序列,求最大的连续子序列之积。(有正数,负数,0) 思路:正确分析这三种数。0把不同的可能为答案的子序列给隔开了,所以其实可以以0为分隔线将他们拆成多个序列来进行求积,这样就没有0了。
所以LeetCode第53题Maximum Subarray,只需要定义一个变量,用来记录和;本题要定义两个变量:positive和negative,分别记录当前乘积最大值和最小值。 publicintmaxProduct(int[]nums){int max=nums[0];int positive=nums[0];int negative=nums[0];for(int i=1;i<nums.length;i++){positive*=nums[i];negative...
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 ...
149 Max Points on a Line Hard Solution 150 Evaluate Reverse Polish Notation Medium Solution 151 Reverse Words in a String Medium Solution 152 Maximum Product Subarray Medium Solution 153 Find Minimum in Rotated Sorted Array Medium Solution 154 Find Minimum in Rotated Sorted Array II Hard Solution...
(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(...