3 解题代码 ## LeetCode 496fromtypingimportListclassSolution:defnextGreaterElement(self,nums1:List[int],nums2:List[int])->List[int]:res={}stack=[]foriinnums2[::-1]:## 从 num2 最右边的元素开始遍历whilestackandi>=stack[-1]:## 如果栈非空,且该元素比栈顶(列表最右边)元素大stack.pop()...
Explanation: For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1. For number 1 in the first array, the next greater number for it in the second array is 3. For number 2 in the first array, there is no next greater n...
若栈空,则此时cur对应位置之后,不存在较大数 若栈非空,则此时栈顶数即为cur对应位置后最近较大数 将以上结果存入字典hashTable中,便于生成结果 nums2遍历结束后,遍历nums1 以当前遍历数,向hashTable中取值,并存入结果列表res 最后返回res 代码如下: class Solution(object): def nextGreaterElement(self,nums1,n...
我的python代码: 1classSolution(object):2defnextGreaterElement(self, findNums, nums):3"""4:type findNums: List[int]5:type nums: List[int]6:rtype: List[int]7"""8res =[]9foriinfindNums:10index =nums.index(i)11index2 = -112forjinrange(index + 1, len(nums)):13ifnums[j] >i:...
The stack will first contain [9, 8, 7, 3, 2, 1] and then we see 6 which is greater than 1 so we pop 1 2 3 whose next greater element should be 6 解法:栈,递减栈。先求出nums2中所有元素的右边第一个较大数字的位置,并记录到map中。然后,因为nums1是子数组,循环nums1中的元素,记录在...
publicintnextGreaterElement(int n){String value=String.valueOf(n);char[]digits=value.toCharArray();int i=digits.length-1;//找到小于右侧任意值的第一个正整数while(i>0){if(digits[i-1]<digits[i]){break;}i--;}if(i==0){return-1;}//找到该整数右侧大于该整数的最小整数int maxIndex=i,...
Leetcode 556. Next Greater Element III 2. Solution **解析:**Version 1,先将数字n变为字符数组,要找最小的大于n的数,则应该从右往左开始,依次寻找第i位字符右边的大于当前字符的最小数字,然后互换二者位置,由于新数字的第i位字符大于n中的第i位字符,因此新数字i位之后的字符应该从小到大排列,这样可以...
496. Next Greater Element I (without duplicates)nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1’s elements in the ...
Given a circular array (the next element of the last element is the first element of the array), print the Next Greater Number for every element. The Next Greater Number of a number x is the first greater number to its traversing-order next in the array, which means you could search ci...
click (XCUIElement) click (XCUICoordinate) macos: scroll Perform scroll gesture on an element or by relative/absolute coordinates Arguments References scrollByDeltaX:deltaY: (XCUIElement) scrollByDeltaX:deltaY: (XCUICoordinate) macos: rightClick ...