496. The Next Greater Element I
Title description
Give you two arrays nums1 and nums2 with no repeated elements, where nums1 is a subset of nums2.
Please find out the next greater value of each element in nums1 in nums2.
The next greater element of the number x in nums1 refers to the first element greater than x to the right of the corresponding position in nums2. If it does not exist, the corresponding position outputs -1.
Example 1:
Input: nums1 = [4,1,2], nums2 = [1,3,4,2].
Output: [-1,3,-1]
Explanation:
For the number 4 in num1, you cannot enter the second The next higher number is found in the array, so -1 is output.
For the number 1 in num1, the next larger number to the right of the number 1 in the second array is 3.
For the number 2 in num1, there is no next higher number in the second array, so -1 is output.
Example 2:
Input: nums1 = [2,4], nums2 = [1,2,3,4].
Output: [3,-1]
Explanation:
For the number 2 in num1, the next larger number in the second array It is 3.
For the number 4 in num1, there is no next higher number in the second array, so -1 is output.
Problem-solving ideas
Violence law:
For each element in nums1[i], first find it in nums2, and then traverse to the right to find the first element greater than nums1[i].
/**
* 暴力法
* 关键信息：两个数组各自 没有重复元素。
* 模拟题目的意思：对于每一个 nums1[i] 中的元素，先在 nums2 中找到它，然后向右遍历找到第 1 个大于 nums1[i] 的元素。
*
* @param nums1
* @param nums2
* @return
*/
public static int[] nextGreaterElement(int[] nums1, int[] nums2) {
//把数组2的元素放进map中，键为num2的值，值为下标
HashMap<Integer,Integer> map = new HashMap<>();
for (int i = 0; i < nums2.length; i++) {
map.put(nums2[i],i);
}
//定义需要返回的数组
int[] res = new int[nums1.length];
//依次遍历num1中的元素，
for (int i = 0; i < nums1.length; i++) {
for (int j = map.get(nums1[i]); j < nums2.length; j++) {
if (nums2[j] > nums1[i]){
res[i] = nums2[j];
break;
}
else {
res[i] = -1;
}
}
}
return res;
}
Monotonic stack
/**
* 单调栈
* @param nums1
* @param nums2
* @return
*/
public static int[] nextGreaterElement_1(int[] nums1, int[] nums2) {
int len1 = nums1.length;
int len2 = nums2.length;
Deque<Integer> stack = new ArrayDeque<>();
Map<Integer,Integer> map = new HashMap<>();
//先处理nums2，把对应关系存入哈希表
for (int i = 0; i < len2; i++) {
while (!stack.isEmpty() && stack.peekLast() < nums2[i]){
map.put(stack.removeLast(),nums2[i]);
}
stack.addLast(nums2[i]);
}
int[] res = new int[len1];
for (int i = 0; i < len1; i++) {
res[i] = map.getOrDefault(nums1[i],-1);
}
return res;
}