503. Next Greater Element II

Posted 2020-09-27 apanda009

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 circularly to find its next greater number. If it doesn‘t exist, output -1 for this number.

Example 1:

Input: [1,2,1]
Output: [2,-1,2]
Explanation: The first 1‘s next greater number is 2; 
The number 2 can‘t find next greater number; 
The second 1‘s next greater number needs to search circularly, which is also 2.

我们可以使用栈来进行优化上面的算法，我们遍历两倍的数组，然后还是坐标i对n取余，取出数字，如果此时栈不为空，且栈顶元素小于当前数字，说明当前数字就是栈顶元素的右边第一个较大数，那么建立二者的映射，并且去除当前栈顶元素，最后如果i小于n，则把i压入栈。因为res的长度必须是n，超过n的部分我们只是为了给之前栈中的数字找较大值，所以不能压入栈，参见代码如下：

没有用hashMap 存-1的值, 用的是Arrays.fill(nums, -1)

public int[] nextGreaterElements(int[] nums) {
        int n = nums.length, next[] = new int[n];
        Arrays.fill(next, -1);
        Stack<Integer> stack = new Stack<>(); // index stack
        for (int i = 0; i < n * 2; i++) {
            int num = nums[i % n]; 
            while (!stack.isEmpty() && nums[stack.peek()] < num)
                next[stack.pop()] = num;
            if (i < n) stack.push(i);
        }   
        return next;
    }