算法练习Stack(七)--P496

2020-03-07

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:

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

一开始理解错了。是说第一个数组中的对应的值：首先找到这个值在第二个数组中的位置（不是这个值在第一个数组中的位置相等的第二个数组中的位置）假设第一个数组的第二个位置是3；要找的是第二个数组中3的位置，而不是第二个数组的第二个位置。

暴-这里好像有个不那么敞亮的表达-力-的问题-破bl不能使用-解有效，速度比较慢。而且完全没有用到Stack数据结构—— Runtime: 11 ms, faster than 5.88% of Java online submissions for Next Greater Element I.
Memory Usage: 37.2 MB, less than 100.00% of Java online submissions for Next Greater Element I.

public int[] nextGreaterElement(int[] nums1, int[] nums2) {
        int[] f = new int[nums1.length];

        for (int i = 0; i < nums1.length; i++) {
            int t = nums1[i];

            boolean u = false, found = false;
            for (int j = 0; j < nums2.length; j++) {
                int n = nums2[j];
                if(n == t) {
                    u = true;
                }else if(u && n > t) {
                    f[i] = n;
                    found = true;
                    break;
                }
            }
            if(!found) {
                f[i] = -1;
            }
        }
        return f;
    }