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.

轉載註明出處：http://www.cnblogs.com/wdfwolf3/，謝謝。

時間複雜度O(n)，51ms。此題變成循環數組來容許查找nextGreaterElement，因此能夠當作數組nums後面拼接一個nums，這樣對於每一個元素它後面都有一份完整的nums，按照題1的作法處理nums，區別是這裏stack中是索引，不是值。

public int[] nextGreaterElement(int[] nums){
        //ans數組存放結果
        int[] ans = new int[nums.length];
　　　　 //輔助棧，存放待查找結果元素的索引，找到的當即出棧。
        Stack<Integer> stack = new Stack<>();
        //第一次遍歷nums，同第一個問題，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);
        }
        //第二次遍歷nums，至關於在元素的左邊查找nextGreaterElement，可是再也不入棧，由於元素在第一次遍歷時已經被遍歷過，不能再次入棧
        for (int i = 0; i < nums.length; i++) {
            while(!stack.isEmpty() && nums[stack.peek()] < nums[i]){
                ans[stack.peek()] = nums[i];
                stack.pop();
            }
        }
        //此時棧中存放的是沒有找到nextGreaterElement的元素，在結果中賦值-1
        for (Integer i : stack){
            ans[i] = -1;
        }
        return ans;
    }