LeetCode - The World's Leading Online Programming Learning Platformleetcode.com/problems/maximum-gcd-sum-of-a-subarray/description/ 简单介绍一下题目大意,更详细的描述可以见链接中的原题描述及样例数据: 给定一个长度为n的正整数数组nums。对于nums的任意一个子数组(连续、非空的一段),定义s为子数组...
有了FIND-MAX-CROSSING-SUBARRAY我们可以找到跨越中点的最大子数组,于是,我们也可以设计求解最大子数组问题的分治算法了,其伪代码如下: FIND-MAXMIMUM-SUBARRAY(A, low, high): if high = low return (low, high, A[low]) else mid = floor((low+high)/2) (left-low, left-high, left-sum) = FIND...
dp[i] represents the maximum sum of subarray which ends in nums[i], and dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]). and since we have to include nums[i] due to it’s on the defination of dp[i], and when dp[i-1]<0 we can just choose not to add it. so the...
=max(Max(A[i]), sum-sum)=0 含义为:Max subarray为空数组。 显然不满足题意。 因此,在这种情况下,我们要求的最大和,只能是Max(A[i])=allMax 代码参考: 1classSolution {2public:3intmaxSubarraySumCircular(vector<int>&A) {4intsum=A[0];5intcurMax=A[0], curMin=A[0], allMax=A[0], ...
Maximum Size Subarray Sum Equals K Given an arraynumsand a target valuek, find the maximum length of a subarray that sums tok. If there isn't one, return 0 instead. Example 1: Givennums=[1, -1, 5, -2, 3],k=3, return4. (because the subarray[1, -1, 5, -2]sums to 3 and...
a) Maximum subarray sum in left half b) Maximum subarray sum in right half c) Maximum subarray sum such that the subarray crosses the midpoint*/returnmax(max(maxSubArray(arr, l, m),//最左侧数组求最大summaxSubArray(arr, m+1, h)),//对右侧数组求最大sum ,之后求左右的最大值maxCrossing...
p2p2= maximum subarray sum in[l…r][l…r] p1p1= remaining left side sum p3p3= remaining right side sum p1+p2+p3=∑a[l…r]p1+p2+p3=∑a[l…r] Now to combine two arrays, the maximum subarray sum of the combined array would bemax(p(1)2,p(2)2,p(1)2+p(1)3+p(2)1+p(...
如果已知A[:i-1]的子序列最大和sum(i-1),那A[:i]的解就是取sum(i-1)和i位置的局部最大值的较大值。那么如何求i位置的局部最大值,同样也是一个DP问题,对i-1位置的局部最大值进行更新。 此题对于没有DP思想的coder来说其实是很难的一道题,但是LeetCode将其分类为easy。
有了FIND-MAX-CROSSING-SUBARRAY我们可以找到跨越中点的最大子数组,于是,我们也可以设计求解最大子数组问题的分治算法了,其伪代码如下: FIND-MAXMIMUM-SUBARRAY(A, low, high): if high = low return (low, high, A[low]) else mid = floor((low+high)/2) (left-low, left-high, left-sum)...
To solve the problem of finding the maximum circular subarray sum in a given array, the program needs to handle two scenarios: the maximum sum subarray that does not wrap around and the maximum sum subarray that does wrap around the array. The solution involves using Kadane's algorithm to fi...