leetcode503—— Next Greater Element II （JAVA)

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.

Note: The length of given array won't exceed 10000.

转载注明出处：http://www.cnblogs.com/wdfwolf3/，谢谢。

时间复杂度O(n)，51ms。此题变成循环数组来容许查找nextGreaterElement，因此能够当作数组nums后面拼接一个nums，这样对于每一个元素它后面都有一份完整的nums，按照题1的作法处理nums，区别是这里stack中是索引，不是值。

public int[] nextGreaterElement(int[] nums){
        //ans数组存放结果
        int[] ans = new int[nums.length];
　　　　 //辅助栈，存放待查找结果元素的索引，找到的当即出栈。
        Stack<Integer> stack = new Stack<>();
        //第一次遍历nums，同第一个问题，查找元素的nextGreaterElement。
        for (int i = 0; i < nums.length; i++) {
            while(!stack.isEmpty() && nums[stack.peek()] < nums[i]){
                ans[stack.peek()] = nums[i];
                stack.pop();
            }
            stack.push(i);
        }
        //第二次遍历nums，至关于在元素的左边查找nextGreaterElement，可是再也不入栈，由于元素在第一次遍历时已经被遍历过，不能再次入栈
        for (int i = 0; i < nums.length; i++) {
            while(!stack.isEmpty() && nums[stack.peek()] < nums[i]){
                ans[stack.peek()] = nums[i];
                stack.pop();
            }
        }
        //此时栈中存放的是没有找到nextGreaterElement的元素，在结果中赋值-1
        for (Integer i : stack){
            ans[i] = -1;
        }
        return ans;
    }