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L503.java
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package com.liang.leetcode.stack;
import java.util.Arrays;
import java.util.Stack;
/**
* @ClassName L503
* @description 下一个更大元素 II
* @Author LiaNg
* @Date 2019-11-25
*/
public class L503 {
public static void main(String[] args) {
int[] nums = new int[]{1, 2, 1};
L503 l503 = new L503();
System.out.println("l503.nextGreaterElements(nums) = " + Arrays.toString(l503.nextGreaterElements(nums)));
}
/**
* 给定一个循环数组（最后一个元素的下一个元素是数组的第一个元素），输出每个元素的下一个更大元素。数字 x 的下一个更大的元素是按数组遍历顺序，这个数字之后的第一个比它更大的数，这意味着你应该循环地搜索它的下一个更大的数。如果不存在，则输出
* -1。
* 来源：力扣（LeetCode）
* 链接：https://leetcode-cn.com/problems/next-greater-element-ii
* 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
*/
public int[] nextGreaterElements(int[] nums) {
int[] result = new int[nums.length];
Stack<Integer[]> stack = new Stack<>();
Arrays.fill(result, -1);
for (int i = 0; i < nums.length * 2; i++) {
while (!stack.empty() && nums[i % nums.length] > stack.peek()[1]) {
result[stack.pop()[0]] = nums[i % nums.length];
}
stack.push(new Integer[]{i % nums.length, nums[i % nums.length]});
}
return result;
}
/**
* LeetCode 耗时最短解答
*/
public int[] nextGreaterElements1(int[] nums) {
int[] nextBiggerPos = new int[nums.length];
Arrays.fill(nextBiggerPos, -1);
// 先正向
for (int i = nums.length - 2; i >= 0; i--) {
if (nums[i] < nums[i + 1]) {
nextBiggerPos[i] = i + 1;
} else {
int nextPossiblePos = nextBiggerPos[i + 1];
while (true) {
if (nextPossiblePos == -1 || nums[i] < nums[nextPossiblePos]) {
nextBiggerPos[i] = nextPossiblePos;
break;
}
nextPossiblePos = nextBiggerPos[nextPossiblePos];
}
}
}
// 再循环
for (int i = 0; i < nums.length; i++) {
if (nextBiggerPos[i] == -1) {
int nextPossiblePos = 0;
while (nextPossiblePos < i) {
if (nextPossiblePos == -1 || nums[nextPossiblePos] > nums[i]) {
nextBiggerPos[i] = nextPossiblePos;
break;
} else {
nextPossiblePos = nextBiggerPos[nextPossiblePos];
}
}
}
}
int[] output = new int[nums.length];
for (int i = 0, len = nums.length; i < len; i++) {
output[i] = nextBiggerPos[i] == -1 ? -1 : nums[nextBiggerPos[i]];
}
return output;
}
}