Given a circular integer arraynums(i.e., the next element ofnums[nums.length - 1]isnums[0]), returnthe next greater number for every element innums. The next greater number of a numberxis the first greater number to its traversing-order next in the array, which means you could search...
Can you solve this real interview question? Next Greater Numerically Balanced Number - An integer x is numerically balanced if for every digit d in the number x, there are exactly d occurrences of that digit in x. Given an integer n, return the smallest
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 number for it...
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...
如果遍历完成了一次Nums,栈中还有元素,即有些元素没有找到next greater number,按照题目要求,这些元素对应的位置应该设定成0,所以我们可以在res初始化的时候就设定默认值为0. 举例说明,注意栈中保存的是下标: Input: [2,7,4,3,5] Output: [7,0,5,5,0] ...
Output: [-1,3,-1] 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, ...
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
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,...
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,nums2): ...
(int i = nums.size() - 1; i >= 0; i--) { // 判定个子高矮 while (!s.empty() && s.top() <= nums[i]) { // 矮个起开,反正也被挡着了。。。 s.pop(); } // nums[i] 身后的 next great number res[i] = s.empty() ? -1 : s.top(); // s.push(nums[i]); } ...