想到了用哈希表存这个数的位置,但是没有想到可以直接用哈希表存next great,用栈存还没找到的数,没遍历一个数就考察栈中的元素小,小的话,这个数就是栈中数的next great,栈中的数肯定是下大上小。 publicint[] nextGreaterElement(int[] nums1,int[] nums2) {/*通过map建立当前元素和其next great的映射 ...
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,j...
Next Greater Element II 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 m...
Can you solve this real interview question? Next Greater Element I - The next greater element of some element x in an array is the first greater element that is to the right of x in the same array. You are given two distinct 0-indexed integer arrays num
Can you solve this real interview question? Next Greater Element II - Given a circular integer array nums (i.e., the next element of nums[nums.length - 1] is nums[0]), return the next greater number for every element in nums. The next greater number of
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...
def nextGreaterElement(self,nums1,nums2): stk=[] #单调栈 hashTable={} #哈希表 for i in range(len(nums2)-1,-1,-1): #反向遍历 print(i) cur=nums2[i] while stk!=[] and stk[-1]<=cur: #关键 stk.pop() if stk: hashTable[cur]=stk[-1] ...
## 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()## 清空这个...
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 number for it in the second array, so output -1. 1. 2. 3. 4. 5. 6. ...
public int[] nextGreaterElement(int[] nums1, int[] nums2) { int[] res = new int[nums1.length]; // we are looking for the right side closest and larger element for each element in nums1 for (int i = 0; i < nums1.length; i++) { ...