【每天一道编程系列-2018.2.18】（Ans）

【题目描述】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)). 

【题目翻译】两个排序数组，找这两个排序数组的中位数，时间复杂度为O(log(m+n)) 

【解题思路】

采用类二分查找算法 

【本题答案】
/**
* @author yesr
* @create 2018-02-18 下午11:06
* @desc
**/
public class Test0218 {
/**
* <pre>
* 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)).
*
* 题目大意：
* 两个排序数组，找这两个排序数组的中位数，时间复杂度为O(log(m+n))
*
* 题解思路：
* 递归分治求解
* </pre>
*
* @param nums1
* @param nums2
* @return
*/
public double findMedianSortedArrays(int[] nums1, int[] nums2) {

if (nums1 == null) {
nums1 = new int[0];
}

if (nums2 == null) {
nums2 = new int[0];
}

int len1 = nums1.length;
int len2 = nums2.length;

if (len1 < len2) {
// 确保第一个数组比第二个数组长度大
return findMedianSortedArrays(nums2, nums1);
}

// 如果长度小的数组长度为0，就返回前一个数组的中位数
if (len2 == 0) {
return (nums1[(len1 - 1) / 2] + nums1[len1 / 2]) / 2.0;
}

int lo = 0;
int hi = len2 * 2;
int mid1;
int mid2;
double l1;
double l2;
double r1;
double r2;

while (lo <= hi) {
mid2 = (lo + hi) / 2;
mid1 = len1 + len2 - mid2;

l1 = (mid1 == 0) ? Integer.MIN_VALUE : nums1[(mid1 - 1) / 2];
l2 = (mid2 == 0) ? Integer.MIN_VALUE : nums2[(mid2 - 1) / 2];

r1 = (mid1 == len1 * 2) ? Integer.MAX_VALUE : nums1[mid1 / 2];
r2 = (mid2 == len2 * 2) ? Integer.MAX_VALUE : nums2[mid2 / 2];

if (l1 > r2) {
lo = mid2 + 1;
} else if (l2 > r1) {
hi = mid2 - 1;
} else {
return (Math.max(l1, l2) + Math.min(r1, r2)) / 2;
}
}

return -1;
}
}