Problem Difinition: Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array[−2,1,−3,4,−1,2,1,−5,4], the contiguous subarray[4,−1,2,1]has the largest sum =6. More practice: If you have fig...
此时由于题目要求数组连续,所以没法保留原sum,所以只能让sum等于从A[i]开始的新的一段数了,这一段数字形成新的candidate。 3. 如果每次得到新的candidate都和全局的max_sum进行比较,那么必然能找到最大的max sum subarray. 在循环过程中,用max_sum记录历史最大的值。从A[0]到A[n-1]一步一步地进行。 其实...
int maxSubArray(int A[], int n) { int ret=maxsub(A,0,n-1); return ret; } protected: int maxsub(int A[], int start, int end) { int max_left=INT_MIN,max_mid=INT_MIN,max_right=INT_MIN; if(start==end) return A[start]; if(start+1==end) { int a=max(A[start],A[end...
Can you solve this real interview question? Maximum Subarray - Given an integer array nums, find the subarray with the largest sum, and return its sum. Example 1: Input: nums = [-2,1,-3,4,-1,2,1,-5,4] Output: 6 Explanation: The subarray [4,-1,2,1]
Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn’t one, return 0 instead. Hashmap is also used in this problem. Iterate the array and calculate the sum, if map[sum-k] in the hashmap, update the maxlength=max(maxlength...
def maxSubArray(self, nums: List[int]) -> int: """ Epoch2: 最大子数组和 """ n_sum = 0 res = -float('inf') for n in nums: n_sum += n res = max(res, n_sum) if n_sum < 0: n_sum = 0 return res 334. 递增的三元子序列:给你一个整数数组 nums,判断这个数组中是否存在...
public int maxSubArray(int[] nums) { if(nums==null || nums.length == 0) return 0; int[] dp = new int[nums.length]; dp[0] = nums[0]; int res = dp[0]; for(int i = 1; i<nums.length;i++){ dp[i] = nums[i] + (dp[i-1]<0 ? 0: dp[i-1]); ...
0053-maximum-subarray Time: 67 ms (90.80%), Space: 70.2 MB (29.39%) - LeetHub May 30, 2024 0056-merge-intervals Time: 27 ms (33.59%), Space: 22.6 MB (68.49%) - LeetHub May 31, 2024 0058-length-of-last-word Clean up repo Jan 21, 2024 ...
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边界情况的处理, max 初始值设为最小int肯定会被更新, preMin设为0确保了最少包含一个值,刻意应付全部都是负数的corner case。 Code: # code blockpublicintmaxSubArray(int[]nums){if(nums==null||nums.length==0){return-1;}intmax=Integer.MIN_VALUE;intpreSum=0;intpreMin=0;for(inti=0;i<nums.le...