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 O(log (m+n)).
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
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【题目分析】
给定两个升序排列的整数数组,找到这两个数组所有元素的中位数。
【思路】
【java代码】
public class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int[] A = nums1, B = nums2;
int m = nums1.length, n = nums2.length; if(m > n){
A = nums2; B = nums1;
m = nums2.length; n = nums1.length;
} if(n == 0) return 0; int mini = 0, maxi = m;
int i = 0, j = 0;
int maxofleft, minofright = 0; while(mini <= maxi){
i = (maxi - mini)/2 + mini;
j = (m + n + 1)/2 - i;
if(j > 0 && i < m && B[j-1] > A[i]){
mini = i + 1;
} //i is too small
else if(i > 0 && j < n && A[i-1] > B[j]){
maxi = i - 1;
} //i is too big
else{
if(i == 0) maxofleft = B[j-1];
else if(j == 0) maxofleft = A[i-1];
else maxofleft = Math.max(A[i-1], B[j-1]); if((m + n)%2 == 1) return maxofleft; if(i == m) minofright = B[j];
else if(j == n) minofright = A[i];
else minofright = Math.min(B[j], A[i]); return (double)(maxofleft + minofright) / 2.0;
}
}
return 0;
}
}