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]