metge sort的merge时候比较,o(n+m)
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public class Solution { public static void main(String[] args) { // TODO Auto-generated method stub int A[] = {}; int B[] = { 1, 2, 3, 4, 5 }; double c = new Solution().findMedianSortedArrays(A, B); System.out.println(c); } public double findMedianSortedArrays(int A[], int B[]) { // Start typing your Java solution below // DO NOT write main() function int startA = 0; int startB = 0; int tempS = 0, tempE = 0; double end = ((double) A.length + B.length) / 2; while (startA + startB < end) { if (startA >= A.length) { tempS = B[startB]; startB++; } else if (startB >= B.length) { tempS = A[startA]; startA++; } else if (A[startA] > B[startB]) { tempS = B[startB]; startB++; } else { tempS = A[startA]; startA++; } } if ((A.length + B.length) % 2 == 1) { tempE = tempS; } else if (startA >= A.length) { tempE = B[startB]; } else if (startB >= B.length) { tempE = A[startA]; } else if (A[startA] > B[startB]) { tempE = B[startB]; } else { tempE = A[startA]; } return ((double) tempS + tempE) / 2; } }
改进:
比较两个数组的中位数
ar1[]和ar2[]为输入的数组
算法过程:
1.得到数组ar1和ar2的中位数m1和m2
2.如果m1==m2,则完成,返回m1或者m2
3.如果m1>m2,则中位数在下面两个子数组中
a) From first element of ar1 to m1 (ar1[0...|_n/2_|])
b) From m2 to last element of ar2 (ar2[|_n/2_|...n-1])
4.如果m1<m2,则中位数在下面两个子数组中
a) From m1 to last element of ar1 (ar1[|_n/2_|...n-1])
b) From first element of ar2 to m2 (ar2[0...|_n/2_|])
5.重复上面的过程,直到两个子数组的大小都变成2
6.如果两个子数组的大小都变成2,使用下面的式子得到中位数
Median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2
时间复杂度:O(logn)。
待实现代码。。。