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:
nums1 and nums2 are unique.nums1 and nums2 would not exceed 1000.
package leetcode.easy;
public class NextGreaterElementI {
private static void print_arr(int[] nums) {
for (int num : nums) {
System.out.print(num + " ");
}
System.out.println();
}
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
int[] arr = new int[nums1.length];
for (int i = 0; i < nums1.length; i++) {
int current = nums1[i];
int j = 0;
while (nums2[j] != current) {
j++;
}
for (j = j + 1; j < nums2.length; j++) {
if (nums2[j] > current) {
arr[i] = nums2[j];
break;
}
}
if (j == nums2.length) {
arr[i] = -1;
}
}
return arr;
}
@org.junit.Test
public void test1() {
int[] nums1 = { 4, 1, 2 };
int[] nums2 = { 1, 3, 4, 2 };
print_arr(nextGreaterElement(nums1, nums2));
}
@org.junit.Test
public void test2() {
int[] nums1 = { 2, 4 };
int[] nums2 = { 1, 2, 3, 4 };
print_arr(nextGreaterElement(nums1, nums2));
}
}
LeetCode_496. Next Greater Element I
原文：https://www.cnblogs.com/denggelin/p/12117373.html