Median of Two Sorted Arrays (全数组扫描)

最新推荐文章于 2021-12-03 22:06:10 发布

原创最新推荐文章于 2021-12-03 22:06:10 发布 · 671 阅读

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本文介绍了一个算法，用于找到两个已排序数组的中位数，该算法通过比较数组元素并进行适当的调整来实现。适用于不同长度的数组，并能够高效地找到中位数。

public class Solution {

public double findMedianSortedArrays(int A[], int B[]) {
// Start typing your Java solution below
// DO NOT write main() function
int alength = A.length;
int blength = B.length;

if (alength == 0 ) {
if (blength % 2 == 1) {
return B[blength/2];
}
else {
return ( B[blength/2 - 1] + B[blength/2] ) / 2.0;
}
}
else if (blength == 0 ) {
if (alength % 2 == 1) {
return A[alength/2];
}
else {
return ( A[alength/2 - 1] + A[alength/2] ) / 2.0;
}
}

int left = 0;
int right = alength-1;

return findMedian(A, B, left, right, alength, blength);

}

public double findMedian(int A[], int B[], int l, int r, int nA, int nB) {
if (l>r) {
int left = 0;
int right = nB-1;
return findMedian(B, A, left, right, nB, nA);
}


int i = (l+r)/2;
int j = (nA+nB)/2-i-1;

boolean lvalid = false;
if (j <= nB - 1 ) {
lvalid = (j < 0) || (A[i] >= B[j]);
}
else {
lvalid = (j < 0);
}

boolean rvalid = false;
if (j+1 >=0 ) {
rvalid = (j >= nB - 1) || (A[i] <= B[j + 1]);
}
else {
rvalid = (j >= nB - 1);
}

if ( lvalid && !rvalid ) return findMedian(A, B, l, i-1, nA, nB);
else if (!lvalid && rvalid) return findMedian(A, B, i+1, r, nA, nB);
else {
if ( (nA+nB)%2 == 1 ) return A[i];
else if (i>0) {
if (j<0) {
return (A[i]+A[i-1])/2.0;
}
else {
return (A[i]+Math.max(B[j], A[i-1]))/2.0;
}

}
else return (A[i]+B[j])/2.0;
}
}

}