...2.选择子数组(Select Subarray):根据分解步骤中得到的子数组和枢纽元素的位置,确定要继续查找的子数组。...如果 K 大元素的位置在枢纽元素的右侧,那么在右侧的子数组中继续查找;如果在左侧,那么在左侧的子数组中查找。3.递归(Recursion):递归地在所选子数组中查找第 K 大元素。...这个过程会反复进行,直到...
while len(result) < len(left) + len(right): if left[index_left] <= right[index_right]: result.append(left[index_left]) index_right += 1 else: result.append(right[index_right]) index_left += 1 if index_right == len(right): result += left[index_left:] break if index_left =...
def findLongestSubarrayBySum(s, arr): """Return a two-element array that represent the bounds of the subarray.""" ans = [-1, -2] s -= arr[0] right = 0 for left in range(len(arr)): while right < len(arr) - 1 and s >= arr[right + 1]: s -= arr[right + 1] right...
Start_index:Start index of theSUBARRAYto be chosen from parent 1End_index: End index of theSUBARRAYto be chosen from parent 1Return Value: Return child after performing the crossover (also a list containing a valid sequence of cities) Function Definition:In this function, students are asked t...
The algorithm starts with key_item = 2 and goes through the subarray to its left to find the correct position for it. In this case, the subarray is [8].Since 2 < 8, the algorithm shifts element 8 one position to its right. The resultant array at this point is [8, 8, 6, 4, ...
(result)<len(left)+len(right):ifleft[index_left]<=right[index_right]:result.append(left[index_left])index_right+=1else:result.append(right[index_right])index_left+=1ifindex_right==len(right):result+=left[index_left:]breakifindex_left==len(left):result+=right[index_right:]breakreturn...
链接:https://leetcode-cn.com/problems/maximum-subarray/ let nums = [-2,1,-3,4,-1,2,1,-5,4] var maxSubArray = function(nums) { let max = nums[0]; let allmax = nums[0]; for (let i = 1 , n = nums.length; i < n ; i++) { ...
index = i return(index) 5.Maximum Subarray Find the contiguous subarray within anarray(containing at least one number)which has the largest sum. 在数组中找出一个连续的子数组,该子数组至少包含一个数字,并且其所有元素之和最大。 示例: 给定数组[-2, 1,-3, 4, -1, 2, 1, -5, 4],得出子...
classSolution(object):defdominantIndex(self, nums):""":type nums: List[int] :rtype: int"""#思路:对原数组取出第一个最大值,去除该最大值的数组找出第二最大*2,比较importcopyiflen(nums)<=1:return0 max_number=max(nums) nums_=copy.copy(nums) ...
classSolution:defminSubArrayLen(self, s, nums):""":type s: int :type nums: List[int] :rtype: int"""l=0 r=0 sum_all=0 nums_len=len(nums) minLength= nums_len + 1whilel <nums_len:ifr < nums_lenandsum_all < s://如果右指针未出界(<nums_len)且窗口内元素之和尚且小于s ...