496- next higher element Ⅰ
Given two no duplicate elements of the array nums1
and nums2
which nums1
are nums2
a subset. Find nums1
each element in the nums2
case of a value greater than that.
nums1
The next higher element refers to the number x is x in nums2
the first large element ratio x of the right side in the corresponding position. If not, the position output corresponding to -1.
Example 1:
输入: nums1 = [4,1,2], nums2 = [1,3,4,2].
输出: [-1,3,-1]
解释:
对于num1中的数字4,你无法在第二个数组中找到下一个更大的数字,因此输出 -1。
对于num1中的数字1,第二个数组中数字1右边的下一个较大数字是 3。
对于num1中的数字2,第二个数组中没有下一个更大的数字,因此输出 -1。
Example 2:
输入: nums1 = [2,4], nums2 = [1,2,3,4].
输出: [3,-1]
解释:
对于num1中的数字2,第二个数组中的下一个较大数字是3。
对于num1中的数字4,第二个数组中没有下一个更大的数字,因此输出 -1。
note:
nums1
Andnums2
all elements are unique.nums1
Andnums2
the array size is not more than 1,000.
Source: stay button (LeetCode)
link: https://leetcode-cn.com/problems/next-greater-element-i
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
int[] res = new int[nums1.length];
int i;
for (int j = 0; j < nums1.length; j++) {
boolean flag = false;
for (i = 0; i < nums2.length; i++) {
if (nums2[i] == nums1[j]) {
flag = true;
}
if (flag && nums2[i] > nums1[j]) {
res[j] = nums2[i];
break;
}
}
if(flag && i == nums2.length) {
res[j] = -1;
}
}
return res;
}
public int[] nextGreaterElement(int[] nums1, int[] nums2) {
// 单调栈
Stack<Integer> stack = new Stack<>();
HashMap<Integer, Integer> map = new HashMap<>();
for (int value : nums2) {
while (!stack.empty() && value > stack.peek()) {
map.put(stack.pop(), value);
}
stack.push(value);
}
int[] res = new int[nums1.length];
while (!stack.empty()) {
map.put(stack.pop(), -1);
}
for (int j = 0; j < nums1.length; j++) {
res[j] = map.get(nums1[j]);
}
return res;
}