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
一个递归的思想，时间复杂度O(log(m+n))
每次二分两个数组，当a[i-1] > b[j-1]的时候，说明第k大的数字在a的左边和b的右边；否则，第k大的数字在a的右边b的左边。
就这样递归下去，直到求出值。
class Solution { public: double findMedianSortedArrays(int a[], int m, int b[], int n) { int l = (m + n + 1) >> 1; int r = (m + n + 2) >> 1; return (solve(a, m, b, n, l) + solve(a, m, b, n, r)) / 2.0; } int solve(int a[], int m, int b[], int n, int k) { if (m > n) return solve(b, n, a, m, k); if (m == 0) return b[k-1]; if (k == 1) return min(a[0], b[0]); int i = min(m, k / 2), j = min(n, k / 2); if (a[i - 1] > b[j - 1]) return solve(a, m, b + j, n - j, k - j); else return solve(a + i, m - i, b, n, k - i); } };