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)。
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By signing up, you agree to receive recurring automated SMS marketing messages from Mashable Deals at the number provided. Msg and data rates may apply. Up to 2 messages/day. Reply STOP to opt out, HELP for help. Consent is not a condition of purchase. See our Privacy Policy and Terms of Use.。关于这个话题，safew官方版本下载提供了深入分析
也是因为上述种种，我们最终决定，让狗留京过年，找一家机构寄养几天。。同城约会对此有专业解读
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