leetcode合并数组中位数

合并数组求中位数

class Solution {
public:
	double findMedianBaseCase(int med, int C[], int n) {
		if (n == 1)
			return (med+C[0])/2.0;
		if (n % 2 == 0) {
			int a = C[n/2 - 1], b = C[n/2];
			if (med <= a)
				return a;
			else if (med <= b)
				return med;
			else /* med > b */
				return b;
		} else {
			int a = C[n/2 - 1], b = C[n/2], c = C[n/2 + 1];
			if (med <= a)
				return (a+b) / 2.0;
			else if (med <= c)
				return (med+b) / 2.0;
			else /* med > c */
				return (b+c) / 2.0;
		}
	}
	double findMedianBaseCase2(int med1, int med2, int C[], int n) {
		if (n % 2 == 0) {
			int a = (((n/2-2) >= 0) ? C[n/2 - 2] : INT_MIN);
			int b = C[n/2 - 1], c = C[n/2];
			int d = (((n/2 + 1) <= n-1) ? C[n/2 + 1] : INT_MAX);
			if (med2 <= b)
				return (b+max(med2,a)) / 2.0;
			else if (med1 <= b)
				return (b+min(med2,c)) / 2.0;
			else if (med1 >= c)
				return (c+min(med1,d)) / 2.0;
			else if (med2 >= c)
				return (c+max(med1,b)) / 2.0;
			else  /* a < med1 <= med2 < b */
				return (med1+med2) / 2.0;
		} else {
			int a = C[n/2 - 1], b = C[n/2], c = C[n/2 + 1];
			if (med1 >= b)
				return min(med1, c);
			else if (med2 <= b)
				return max(med2, a);
			else  /* med1 < b < med2 */
				return b;
		}
	}
	double findMedianSingleArray(int A[], int n) {
		assert(n > 0);
		return ((n%2 == 1) ? A[n/2] : (A[n/2-1]+A[n/2])/2.0);
	}
	double findMedianSortedArrays(int A[], int m, int B[], int n) {
		assert(m+n >= 1);
		if (m == 0)
			return findMedianSingleArray(B, n);
		else if (n == 0)
			return findMedianSingleArray(A, m);
		else if (m == 1)
			return findMedianBaseCase(A[0], B, n);
		else if (n == 1)
			return findMedianBaseCase(B[0], A, m);
		else if (m == 2)
			return findMedianBaseCase2(A[0], A[1], B, n);
		else if (n == 2)
			return findMedianBaseCase2(B[0], B[1], A, m);
		int i = m/2, j = n/2, k;
		if (A[i] <= B[j]) {
			k = ((m%2 == 0) ? min(i-1, n-j-1) : min(i, n-j-1));
			assert(k > 0);
			return findMedianSortedArrays(A+k, m-k, B, n-k);
		} else {
			k = ((n%2 == 0) ? min(m-i-1, j-1) : min(m-i-1, j));
			assert(k > 0);
			return findMedianSortedArrays(A, m-k, B+k, n-k);
		}
	}
};
原文地址:https://www.cnblogs.com/tgkx1054/p/3086059.html