496. Next Greater Element I

You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1's elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
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, there is no next greater number for it in the second array, so output -1.

 

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
    For number 2 in the first array, the next greater number for it in the second array is 3.
    For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

 

Note:

1. All elements in nums1 and nums2 are unique.
2. The length of both nums1 and nums2 would not exceed 1000.

http://www.cnblogs.com/grandyang/p/6399855.html

 

Key observation:
Suppose we have a decreasing sequence followed by a greater number
For example [5, 4, 3, 2, 1, 6] then the greater number 6 is the next greater element for all previous numbers in the sequence

We use a stack to keep a decreasing sub-sequence, whenever we see a number x greater than stack.peek() we pop all elements less than x and for all the popped ones, their next greater element is x
For example [9, 8, 7, 3, 2, 1, 6]
The stack will first contain [9, 8, 7, 3, 2, 1] and then we see 6 which is greater than 1 so we pop 1 2 3 whose next greater element should be 6

用栈对遍历过得元素先进行条件存储 或条件pop, 对pop出来的元素进行处理, 这样就不用多次遍历, 省的力气在于一次条件pop能够找到遍历过得 元素的题目要求值.在于题目条件时要求大于的, 

stack.peek() < num 让栈内元素有了顺序
hashmap 存储的键值对, 值为结果值, 加上getOrDefault 方法

public int[] nextGreaterElement(int[] findNums, int[] nums) {
        Map<Integer, Integer> map = new HashMap<>(); // map from x to next greater element of x
        Stack<Integer> stack = new Stack<>();
        for (int num : nums) {
            while (!stack.isEmpty() && stack.peek() < num)
                map.put(stack.pop(), num);
            stack.push(num);
        }   
        for (int i = 0; i < findNums.length; i++)
            findNums[i] = map.getOrDefault(findNums[i], -1);
        return findNums;
    }

找数组元素相对顺序的最值, 用栈, 找到后盖存储的存储, 该更新状态的更新状态

正序遍历考虑的是栈栈内的元素跟当前元素比, pop出来的都是已经找到符合题意的,

所以栈内元素是正序逆序, 看跟谁比, 看如何pop出来使得元素符合题意.

一看到不能排序, 用到相对顺序就考虑如何用栈吧