过了但是瞎蒙
\u8BA1\u7B97\u4E2D\u95F4\u957F\u65B9\u5F62\u7684\u957F\u5BBD\u7528\u7C7B\u4F3CVenn\u56FE\u7684\u8BA1\u7B97\uFF1Fmax()\u66F4\u65B0(\u5206\u60C5\u51B5)\u5BF9\u79F
\u8BA1\u7B97\u4E2D\u95F4\u957F\u65B9\u5F62\u7684\u957F\u5BBD\u7528\u7C7B\u4F3CVenn\u56FE\u7684\u8BA1\u7B97\uFF1Fmax()\u66F4\u65B0(\u5206\u60C5\u51B5)\u5BF9\u79F0\u778E\u8499\u3002\n```\nclass Solution {\npublic:\n    long long largestSquareArea(vector<vector<int>>& bottomLeft, vector<vector<int>>& topRight) {\n        long long ans = 0;\n        for(int i = 0; i < bottomLeft.size() - 1; i ++)\n        {\n            for(int j = i + 1; j <  bottomLeft.size(); j ++)\n            {\n                int x1 = bottomLeft[i][0];\n                int y1 = bottomLeft[i][1];\n                int x2 = topRight[i][0];\n                int y2 = topRight[i][1];\n\n                int X1 = bottomLeft[j][0];\n                int Y1 = bottomLeft[j][1];\n                int X2 = topRight[j][0];\n                int Y2 = topRight[j][1];\n                if(\n                    (x1 >= X2) || (y1 >= Y2)\n                    || (x2 <= X1) || (y2 <= Y1)\n                )continue;\n                int minx = max((X2 - x1), (x2 - x1));\n                minx = max(minx, X2 - X1);\n                minx = max(minx, x2 - X1);\n                int miny = max((Y2 - y1), ((y2 - y1)));\n                miny = max(miny, (Y2 - Y1));\n                miny = max(miny, (y2 - Y1));\n                int y = (Y2 - Y1) + (y2 - y1) - miny;\n                int x = (X2 - X1) + (x2 - x1) - minx;\n                int s = min(x, y);\n                long long a = (long long) s * s;\n                ans = max(ans, a);       \n            }              \n        }\n       return ans;\n    }\n};\n```
['C++']