subarray right_sum = float("-Inf") sum = 0 for j in range(mid + 1, high + 1): sum += A[j] if (sum > right_sum): right_sum = sum max_right = j return max_left, max_right, left_sum + right_sum # using divide and conquer to solve maximum subarray problem # time ...
关键点:Divide and Conquer 1publicclassSolution2{3publicintmaxSubArray(int[] nums)4{5//Solution 3: Divide and Conquer. O(nlogn)6if(nums ==null|| nums.length == 0)7return0;8910returnMax_Subarray_Sum(nums, 0, nums.length-1);11}1213publicintMax_Subarray_Sum(int[] nums,intleft,intrigh...
maxSubArray(A, i) = maxSubArray(A, i - 1) >0 ? maxSubArray(A, i - 1) :0 + A[i]; 即得到了n个数的数列与n-1个数的数列的关系,最开始我觉得似乎没有涵盖所有情况,但是后面推导发现所有情况都被这个式子给解决了,很神奇,但是给我自己思考,我确实没有找到思考出这个思想的切入点,实际上它这个...
publicintmaxSubArray(int[]nums){intn=nums.length;//两个变量即可int[]dp=newint[2];intmax=nums[0];dp[0]=nums[0];for(inti=1;i<n;i++){//利用求余,轮换两个变量if(dp[(i-1)%2]<0){dp[i%2]=nums[i];}else{dp[i%2]=dp[(i-1)%2]+nums[i];}max=Math.max(max,dp[i%2])...
packageleetcode// 解法一 DPfuncmaxSubArray(nums[]int)int{iflen(nums)==0{return0}iflen(nums)==1{returnnums[0]}dp,res:=make([]int,len(nums)),nums[0]dp[0]=nums[0]fori:=1;i<len(nums);i++{ifdp[i-1]>0{dp[i]=nums[i]+dp[i-1]}else{dp[i]=nums[i]}res=max(res,dp[i...
the contiguous subarray [4,−1,2,1] has the largest sum = 6. More practice: If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle. Hide Tags Divide and Conquer Array Dynamic Programming ...
classSolution{public:intmaxSubArray(vector<int>&nums){if(nums.size()==0)return0;returnhelper(nums,0,nums.size()-1);}inthelper(vector<int>&nums,int l,int r){if(l>r)returnINT_MIN;//注意此处不是返回0,比如{-2,-1},分治以后变为左中右n{},-1,{-2}三部分。左半部分{}应返回INT_MIN...
,the contiguous subarray [4,-1,2,1] has the largest sum = 6 More practice:If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.分析题目要求O(n)的算法,然而dp对我来说就是老大难,想了半天的我看到题目属于Easy的...
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle. 难度:简单 class Solution { public: int maxSubArray(vector<int>& nums) { int exmax=0; int max=nums[0]; for(int i=0;i<nums.size();i++) { if(ex...
Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum. Example: Input: [-2,1,-3,4,-1,2,1,-5,4], Output: 6 Expla...