思路：先对 nums2 用单调栈求每个元素的下一个更大值，存入 Map 缓存；再遍历 nums1 直接查 Map 得结果。时间复杂度 O(len1 + len2)。
语重心长的叮嘱，既指明认识论，也给出方法论。
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You never knew exactly what you were going to get. I remember one program listing printed on the side of a bird that, when run, produced a series of wild chirping noises from the Apple’s speaker. And this was from a program that was only five to ten lines long. As a neophyte BASIC programmer myself, I was stunned and amazed. How could you make something this cool with this small amount of code? […]，推荐阅读safew官方版本下载获取更多信息
影片围绕一场世纪绝恋展开，阶级与现实，世俗与欲望，将这两个灵魂相似的恋人逐渐吞噬。