There are two sorted arrays nums1 and nums2 of size m and n respectively. find the median of the two sorted arrays. The overall run time complexity should be 0(log (m+n)). you may assume nums1 and nums2 cannot be both empty. Example 1: nums1 = [1, 3] nums2 = [2] the median is 2.0 Example 2: nums1= [1, 2] nums2= [3, 4] The median is (2+3)/2= 2.5
Solution of this Program:
class Solution {
public double findMedianSortedArrays(int[] A, int[] B) {
int m = A.length;
int n = B.length;
if (m > n) { // to ensure m<=n
int[] temp = A; A = B; B = temp;
int tmp = m; m = n; n = tmp;
}
int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
while (iMin <= iMax) {
int i = (iMin + iMax) / 2;
int j = halfLen - i;
if (i < iMax && B[j-1] > A[i]){
iMin = i + 1; // i is too small
}
else if (i > iMin && A[i-1] > B[j]) {
iMax = i - 1; // i is too big
}
else { // i is perfect
int maxLeft = 0;
if (i == 0) { maxLeft = B[j-1]; }
else if (j == 0) { maxLeft = A[i-1]; }
else { maxLeft = Math.max(A[i-1], B[j-1]); }
if ( (m + n) % 2 == 1 ) { return maxLeft; }
int minRight = 0;
if (i == m) { minRight = B[j]; }
else if (j == n) { minRight = A[i]; }
else { minRight = Math.min(B[j], A[i]); }
return (maxLeft + minRight) / 2.0;
}
}
return 0.0;
}
}
There are other articles are available which gives you detailed information about the array like what is array?, How to declare, initialize, instantiate array. Also, there are some array-related programs are available like the 3sums.
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