单调栈
单调栈里存的是数组nums的下标，且栈是单调递增的。所以在遍历数组的过程中，要循环查看栈顶元素对应值是否小于当前值，若小于，则当前元素就是栈顶下标对应位置的“更大的数”，需要弹栈，然后给res赋值记录下来。最后让当前位置下标入栈即可。
这里有个细节，因为要找到整个数组的“更大的数”，即下一个数，所以需要遍历两遍数组。为了代码的统一性，所以循环 2 * n - 1次，下标 采取 i % n 的方式达到循环不越界的效果。
代码如下：
class Solution {
public int[] nextGreaterElements(int[] nums) {
int n = nums.length;
int[] res = new int[n];
Arrays.fill(res, -1);
Deque<Integer> stack = new LinkedList<Integer>();
for (int i = 0; i < 2 * n - 1; i++) {
while (!stack.isEmpty() && nums[stack.peek()] < nums[i % n]) {
res[stack.pop()] = nums[i % n];
}
stack.push(i % n);
}
return res;
}
}
测试代码，便于理解栈的变化
public class Solution {
public int[] nextGreaterElements(int[] nums) {
int n = nums.length;
int[] res = new int[n];
Arrays.fill(res, -1);
Deque<Integer> stack = new LinkedList<Integer>();
for (int i = 0; i < 2 * n - 1; i++) {
System.out.println("i = " + i);
while (!stack.isEmpty() && nums[stack.peek()] < nums[i % n]) {
res[stack.pop()] = nums[i % n];
}
System.out.print("res[]: ");
System.out.println(Arrays.toString(res));
stack.push(i % n);
System.out.print("stack: ");
System.out.println(stack);
}
return res;
}
public static void main(String[] args) {
Solution s = new Solution();
System.out.println(Arrays.toString(s.nextGreaterElements(new int[]{1, 8, -1, -100, -2, -222, 1111111, -111111})));
}
}
i = 0 res[]: [-1, -1, -1, -1, -1, -1, -1, -1] stack: [0] i = 1 res[]: [8, -1, -1, -1, -1, -1, -1, -1] stack: [1] i = 2 res[]: [8, -1, -1, -1, -1, -1, -1, -1] stack: [2, 1] i = 3 res[]: [8, -1, -1, -1, -1, -1, -1, -1] stack: [3, 2, 1] i = 4 res[]: [8, -1, -1, -2, -1, -1, -1, -1] stack: [4, 2, 1] i = 5 res[]: [8, -1, -1, -2, -1, -1, -1, -1] stack: [5, 4, 2, 1] i = 6 res[]: [8, 1111111, 1111111, -2, 1111111, 1111111, -1, -1] stack: [6] i = 7 res[]: [8, 1111111, 1111111, -2, 1111111, 1111111, -1, -1] stack: [7, 6] i = 8 res[]: [8, 1111111, 1111111, -2, 1111111, 1111111, -1, 1] stack: [0, 6] i = 9 res[]: [8, 1111111, 1111111, -2, 1111111, 1111111, -1, 1] stack: [1, 6] i = 10 res[]: [8, 1111111, 1111111, -2, 1111111, 1111111, -1, 1] stack: [2, 1, 6] i = 11 res[]: [8, 1111111, 1111111, -2, 1111111, 1111111, -1, 1] stack: [3, 2, 1, 6] i = 12 res[]: [8, 1111111, 1111111, -2, 1111111, 1111111, -1, 1] stack: [4, 2, 1, 6] i = 13 res[]: [8, 1111111, 1111111, -2, 1111111, 1111111, -1, 1] stack: [5, 4, 2, 1, 6] i = 14 res[]: [8, 1111111, 1111111, -2, 1111111, 1111111, -1, 1] stack: [6, 6] [8, 1111111, 1111111, -2, 1111111, 1111111, -1, 1]
Process finished with exit code 0
