查找循環數組元素下一個較大的值 Next Greater Element II

問題：

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.

解決：

① 數組是一個循環數組，就是說某一個元素的下一個較大值能夠在其前面，那麼對於循環數組的遍歷，爲了使下標不超過數組的長度，咱們須要對n取餘。遍歷每個數字，而後對於每個遍歷到的數字，遍歷全部其餘數字，注意不是遍歷到數組末尾，而是經過循環數組遍歷其前一個數字，遇到較大值則存入結果res中，並break，再進行下一個數字的遍歷。

class Solution { //214ms
    public int[] nextGreaterElements(int[] nums) {
        int len = nums.length;
        int[] res = new int[len];
        Arrays.fill(res,-1);
        for (int i = 0;i < nums.length;i ++){
            for (int j = i + 1;j < i + len;j ++){
                if (nums[j % len] > nums[i]){
                    res[i] = nums[j % len];
                    break;
                }
            }
        }
        return res;
    }
}

② 使用棧來進行優化上面的算法，咱們遍歷兩倍的數組，而後仍是座標i對n取餘，取出數字，若是此時棧不爲空，且棧頂元素小於當前數字，說明當前數字就是棧頂元素的右邊第一個較大數，那麼創建兩者的映射，而且去除當前棧頂元素，最後若是i小於n，則把i壓入棧。由於res的長度必須是n，超過n的部分咱們只是爲了給以前棧中的數字找較大值，因此不能壓入棧。

class Solution { //47ms
    public int[] nextGreaterElements(int[] nums) {
        int len = nums.length;
        int[] res = new int[len];
        Arrays.fill(res,-1);
        Stack<Integer> stack = new Stack<>();
        for (int i = 0;i < 2 * len;i ++){
            int num = nums[i % len];
            while (! stack.isEmpty() && nums[stack.peek()] < num){
                res[stack.peek()] = num;
                stack.pop();
            }
            if (i < len) stack.push(i);
        }
        return res;
    }
}

③ 使用數組實現棧。

class Solution {//26ms     public int[] nextGreaterElements(int[] nums) {         int len = nums.length;         int[] res = new int[len];         Arrays.fill(res,-1);         int[] stack = new int[2 * nums.length];         int top = -1;         for (int i = 0;i < 2 * len;i ++){             int index= i % len;             while(top >= 0 && nums[stack[top]] < nums[index]){                 res[stack[top --]] = nums[index];             }             stack[++ top] = index;         }         return res;     } }