leetcode--Median of Two Sorted Arrays

最新推荐文章于 2023-12-16 16:41:11 发布

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最新推荐文章于 2023-12-16 16:41:11 发布

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原文链接：http://www.cnblogs.com/averillzheng/p/3540115.html

本文介绍了一种在O(log(m+n))时间复杂度下寻找两个有序数组中位数的方法，首先通过合并两个数组并找到中位数，时间复杂度为O(max{m,n}

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There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

method 1, time complexity = O(max{m,n})

public class Solution {
    public double findMedianSortedArrays(int A[], int B[]) {
        double result = 0;
		if(A.length != 0 || B.length != 0){
			if(A.length == 0)
				result = (B[(B.length - 1)/2] + B[B.length/2]) / 2.0;
			if(B.length == 0)
				result = (A[(A.length - 1)/2] + A[A.length/2]) / 2.0;
			int length = A.length + B.length;
			int[] total = new int[length];			
			int index1 = 0;
			int index2 = 0;
			for(int i = 0; i < length / 2 + 1; ++ i){
				if(index1 < A.length && index2 < B.length){
					if(A[index1] <= B[index2]){
						total[i] = A[index1];
						++index1;
					}
					else{
						total[i] = B[index2];
						++index2;
					}
				}
				else if(index1 == A.length){
					total[i] = B[index2];
					++index2;
				}
				else{
					total[i] = A[index1];
					++index1;
				}
			}
			result = (total[(length - 1)/2] + total[length /2]) / 2.0;
		}
		return result;
    }
}

The O(log(m + n)) method

public class Solution {
    public double findMedianSortedArrays(int A[], int B[]) {
       int alen = A.length, blen = B.length;
       if((alen + blen) % 2 != 0) //odd number of elements in total
    	   return findHelper(A, B, (alen + blen) / 2 + 1, 0, alen - 1, 0, blen - 1);
       else //even number of elements in total
    	   return (findHelper(A, B, (alen + blen) / 2, 0, alen - 1, 0, blen - 1) 
    			 + findHelper(A, B, (alen + blen) / 2 + 1, 0, alen - 1, 0, blen - 1)) / 2.0;
    }
	private int findHelper(int A[], int B[], int k, int startA, int endA, int startB, int endB){
		if(endA < startA)
			return B[startB + k - 1];
		if(endB < startB)
			return A[startA + k - 1];
		if(k == 1)
			return Math.min(A[startA], B[startB]);
		
		int lenA = endA - startA + 1;
		int lenB = endB - startB + 1;
		
		if(lenA > lenB)
			return findHelper(B, A, k, startB, endB, startA, endA);
		else {
			int aMid = Math.min(k / 2, lenA);
			int bMid = k - aMid;
			if(A[startA + aMid - 1] < B[startB + bMid - 1])
				return findHelper(A, B, k - aMid, startA + aMid, endA, startB, endB);
			else if(A[startA + aMid - 1] > B[startB + bMid - 1])
				return findHelper(A, B, k - bMid, startA, endA, startB + bMid, endB);
			else //A[startA + aMid - 1] == B[startB + bMid - 1]
				return A[startA + aMid - 1];
		}
    }
}

转载于:https://www.cnblogs.com/averillzheng/p/3540115.html