支撑数查找算法 - Python实现

{"title":"\u5165\u95e8\u3011\u627e\u68c0\u201c\u652f\u6301\u6570\u201d\n\u8d5b\u8d27\u9898\u76ee\n\u65f6\u95f4\u9650\u5236\uff1aC/C++ 1000MS\uff0c\u5176\u4ed6\u8bed\u8a00 2000MS\n\u5185\u5b58\u9650\u5236\uff1aC/C++ 16MB\uff0c\u5176\u4ed6\u8bed\u8a00 32MB\n\u63cf\u8ff0\n\n\u5728\u5df2\u77e5\u4e00\u7ec4\u6574\u6570\u4e2d\uff0c\u6709\u8fd9\u6837\u4e00\u79cd\u6570\u5f88\u5999\uff0c\u5b83\u4eec\u4e0d\u5728\u7b2c\u4e00\u4e2a\uff0c\u4e5f\u4e0d\u5728\u6700\u540e\u4e00\u4e2a\uff0c\u800c\u4e14\u5c31\u597d\u90fd\u6bd4\u5de6\u8fb9\u548c\u53f3\u8fb9\u76f8\u90bb\u7684\u6570\u5927\uff0c\u4f60\u80fd\u627e\u5230\u5b83\u4eec\u5417\uff1f\n\n\u8f93\u5165\u63cf\u8ff0\n\n\u7b2c\u4e00\u884c\u4e3a\u6574\u6570m\uff0c\u8868\u793a\u8f93\u5165\u7684\u6574\u6570\u4e2a\u6570\u3002\uff08 3<= m <=100 )\n\u7b2c\u4e8c\u884c\u4e3a m\u4e2a\u6574\u6570\u3002\n\n\u8f93\u51fa\u63cf\u8ff0\n\n\u8bb8\u591a\u4e2a\u652f\u6301\u6570\uff0c\u6bcf\u884c\u4e00\u4e2a\u3002\n\n\u7528\u4f8b\u8f93\u5165 1 \n\n14\n1 3 2 4 1 5 3 9 7 10 8 23 85 43\n\u7528\u4f8b\u8f93\u51fa 1 \n\n3\n4\n5\n9\n10\n85\n\u5185\u5bb9\uff1a\n\n\u89e3\u9898\u60f3\u8def\uff1a\n\n1. \u901a\u8fc7\u6574\u4e2a\u6570\u7ec4\uff0c\u4ece\u7b2c\u4e8c\u4e2a\u6570\u5230\u53cd\u6570\u7b2c\u4e8c\u4e2a\u6570\uff0c\u5224\u65ad\u8be5\u6570\u662f\u5426\u6bd4\u5176\u5de6\u53f3\u76f8\u90bb\u7684\u6570\u90fd\u5927\u3002\n2. \u5982\u679c\u6ee1\u8db3\u6761\u4ef6\uff0c\u5219\u8f93\u51fa\u8be5\u6570\u3002\n\n\u4ee3\u7801\u5b9e\u73b0\u5982\u4e0b\uff1a\n\npython\nm = int(input())\nnums = list(map(int, input().split()))\n\nfor i in range(1, m-1):\n if nums[i] > nums[i-1] and nums[i] > nums[i+1]:\n print(nums[i])\n\n