Input: [-2,-3,-1] Output: -1 Explanation: Subarray [-1] has maximum sum -1 Note: -30000 <= A[i] <= 30000 1 <= A.length <= 30000 这道题比较好理解,关于最大子数组的解法 可以参考 int maxSubArray(vector<int>& A) { int ans = INT_MIN; int cur = INT_MIN; for (int k ...
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...
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...
同时求的Max subarray和Min subarray的和allMax,allMin 那么,最后的结果则是:res=max(allMax, sum-allMin) ⚠️注意:这里有一个特殊情况:若数列全都是负值, 那么allMax=Max(A[i]),allMin=sum 这时 res=max(allMax, sum-allMin) =max(Max(A[i]), sum-sum)=0 含义为:Max subarray为空数组。 ...
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...
1classSolution:2defmaxSubArray(self, nums: List[int]) ->int:3iflen(nums)<2:returnsum(nums)4res =nums[0] # 最终的子序列和的最大值5temp =[nums[0]] # 最初的子序列为第一个元素6foriinrange(1,len(nums)):7ifnums[i]+sum(temp)<nums[i]: # 关键:如果之前的元素加上当前的元素还比...
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。
(5kyu) The maximum sum subarray problem consists in finding the maximum sum of a contiguous subsequence in an array or list of integers: maxSequence([-2 , 1, -3, 4, -1 , 2, 1, -5, 4]) # should be 6: [4, -1, 2, 1] ...
packageleetcodeimport"math"funcmaxSubarraySumCircular(nums[]int)int{varmax1,max2,sumint// case: no circulationmax1=int(math.Inf(-1))l:=len(nums)fori:=0;i<l;i++{sum+=nums[i]ifsum>max1{max1=sum}ifsum<1{sum=0}}// case: circlingarr_sum:=0fori:=0;i<l;i++{arr_sum+=nums...