索引按值排序，维护索引递增栈。
时间复杂度O(nlogn)
代码如下，顺便说一下，楼主给的答案少了一个..

def nextProgram(l:list):
    l = sorted(list(enumerate(l)), key=lambda x:x[1])
    res = [-1] * len(l)
    stack = []
    for i,e in l:
        while len(stack) > 0 and stack[-1] < i:
            index = stack.pop()
            res[index] = e
        stack.append(i)
    return res

print(nextProgram([11,13,10,5,12,21,3])) 
#[12, 21, 12, 12, 21, -1, -1]