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LeetCode "Median of Two Sorted Arrays"
1A! We get median of each array and compare them, then we know which half should be disguarded and how many should be disguarded.
class Solution {public: double findMidArr(int A[], int s, int e, int &chopLen) { int len = e - s + 1; chopLen = (s + e) / 2 - s; if (len % 2) return A[(s + e) / 2]; else return (A[(s + e) / 2] + A[(s + e) / 2 + 1]) / 2.0; } double findMidMinor(int sm[], int cntSm, int lg[], int cntLg) { vector<int> v; v.assign(lg, lg + cntLg); v.insert(v.end(), sm, sm + cntSm); std::sort(v.begin(), v.end()); size_t len = v.size(); if (len % 2) return v[len / 2]; else return (v[len / 2] + v[len / 2 - 1]) / 2.0; } double findMid(int A[], int sa, int ea, int B[], int sb, int eb) { int lenA = ea - sa + 1; int lenB = eb - sb + 1; // Base cases if (lenA <= 2) return findMidMinor(A + sa, lenA, B + sb, lenB); if (lenB <= 2) return findMidMinor(B + sb, lenB, A + sa, lenA); // Chop int chopLenA, chopLenB; double midA = findMidArr(A, sa, ea, chopLenA); double midB = findMidArr(B, sb, eb, chopLenB); int chopLen = std::min(chopLenA, chopLenB); if (midA == midB) return midA; else if (midA < midB) return findMid(A, sa + chopLen, ea, B, sb, eb - chopLen); else if (midB < midA) return findMid(A, sa, ea - chopLen, B, sb + chopLen, eb); } double findMedianSortedArrays(int A[], int m, int B[], int n) { return findMid(A, 0, m - 1, B, 0, n - 1); }};
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