Median of Two Sorted Arrays

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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大概意思就是两个排序好的数组，如果两个数组合并的话找到其中的中位置数

这是我实现的代码有点复杂

public static double findMedianSortedArrays(int[] nums1, int[] nums2) {int len1 = nums1.length;int len2 = nums2.length;int sum = len1 + len2;int median = sum >> 1;int[] numsTemp = null;int[] numsTempOr = null;int len = 0;boolean lastgt = false;boolean lastlt = false;if (sum % 2 == 0) { // 偶数的时候int index2 = len2 >> 1;int index = 0;numsTemp = nums2;numsTempOr = nums1;len = len1;int medianSub1 = median - 1;for (;;) {if(numsTemp.length > 0 && index2 < numsTemp.length && index2 > -1)index = binarySearch(numsTempOr, numsTemp[index2], len, median - index2);else {numsTemp = nums1;numsTempOr = nums2;index2 = len1 >> 1;len = len2;continue;}if (index + index2 == median) {int index0 = index - 1;if(index2 == 0)return (numsTemp[index2] + numsTempOr[index0]) / 2.0;index = binarySearch(numsTempOr, numsTemp[index2 - 1], len, median - index2);if (index + index2 - 1 == medianSub1) {return (numsTemp[index2] + numsTemp[index2 - 1]) / 2.0;} else {return (numsTemp[index2] + numsTempOr[index0]) / 2.0;}} else if (index + index2 == medianSub1) {int index0 = index;if(index2 + 1 == numsTemp.length)return (numsTemp[index2] + numsTempOr[index]) / 2.0;index = binarySearch(numsTempOr, numsTemp[index2 + 1], len, medianSub1 - index2);if (index + index2 + 1 == median) {return (numsTemp[index2] + numsTemp[index2 + 1]) / 2.0;} else {return (numsTemp[index2] + numsTempOr[index0]) / 2.0;}} else if (index + index2 > median) {if (lastlt == true) {numsTemp = nums1;numsTempOr = nums2;index2 = len1 >> 1;len = len2;lastlt = false;} else {index2--;lastgt = true;}} else {if (lastgt == true) {numsTemp = nums1;numsTempOr = nums2;index2 = len1 >> 1;len = len2;lastgt = false;} else {index2++;lastlt = true;}}}} else { // 奇数的时候int index2 = len2 >> 1;int index = 0;numsTemp = nums2;numsTempOr = nums1;len = len1;for (;;) {if(numsTemp.length > 0 && index2 < numsTemp.length && index2 > -1)index = binarySearch(numsTempOr, numsTemp[index2], len, median - index2);else {numsTemp = nums1;numsTempOr = nums2;index2 = len1 >> 1;len = len2;continue;}if (index + index2 == median) {return numsTemp[index2];} else if (index + index2 > median) {if (lastlt == true) {numsTemp = nums1;numsTempOr = nums2;index2 = len1 >> 1;len = len2;lastlt = false;} else {index2--;lastgt = true;}} else {if (lastgt == true) {numsTemp = nums1;numsTempOr = nums2;index2 = len1 >> 1;len = len2;lastgt = false;} else {index2++;lastlt = true;}}}}}public static int binarySearch(int[] nums, int target, int size, int margin) { // 这个数在数组中前面有多少int begin = 0, end = size;int temp = 0;if(nums.length < 1) {return begin;}for (;;) {temp = (begin + end) >> 1;if (end > begin) {if (target > nums[temp]) {begin = temp + 1;} else if (target == nums[temp]) {if(temp == margin) return margin;while(nums[temp++] == target) {if(temp - 1 == margin) return margin; if(temp == size) {temp ++; break;}}; return --temp;} else {end = temp;}} else {return begin;}}}

这是官方的代码

public static double findMedianSortedArrays2(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;        }        //A是小数组        int iMin = 0, iMax = m, halfLen = (m + n + 1) >> 1;        while (iMin <= iMax) {            int i = (iMin + iMax) >> 1;            int j = halfLen - i;            if (i < iMax && B[j-1] > A[i]){                iMin++; // i is too small, j--, i++            } else if (i > iMin && B[j] < A[i-1]) {                iMax--; // i is too big, j++, i--            } 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;    }

官方的逻辑是这样的：

首先将A数组在随机位置i处分成两部分

First let's cut $\text{A}$ into two parts at a random position $i$:

          left_A             |        right_A    A[0], A[1], ..., A[i-1]  |  A[i], A[i+1], ..., A[m-1]

因为A有m个元素，所以有m+1中分组方式（0~m）

已知:

Since $\text{A}$ has $m$ elements, so there are $m+1$ kinds of cutting ($i=0\sim m$).

And we know:

$\text{len}(\text{left\_A})=i,\text{len}(\text{right\_A})=m-i$.

注意:当i=0时,left_A部分是空数组; i=m时，right_A部分是空数组;

Note: when $i=0$, $\text{left\_A}$ is empty, and when $i=m$, $\text{right\_A}$ is empty.

同理，将B在right_A随机位置j处分成两部分

With the same way, cut $\text{B}$ into two parts at a random position $j$:

          left_B             |        right_B    B[0], B[1], ..., B[j-1]  |  B[j], B[j+1], ..., B[n-1]

将left_A和left_B放入一个区域，同时将right_A和right_B放入另一个区域.并且将这两个区域命名为left_part 和 right_part

Put $\text{left\_A}$ and $\text{left\_B}$ into one set, and put $\text{right\_A}$ and $\text{right\_B}$ into another set. Let's name them $\text{left\_part}$and $\text{right\_part}$:

          left_part          |        right_part    A[0], A[1], ..., A[i-1]  |  A[i], A[i+1], ..., A[m-1]    B[0], B[1], ..., B[j-1]  |  B[j], B[j+1], ..., B[n-1]

当要满足下面两个条件时

If we can ensure:

1. $\text{len}(\text{left\_part})=\text{len}(\text{right\_part})$
2. $\max (\text{left\_part})\le \min (\text{right\_part})$

我们将{A, B}两个数组里面所有的元素分成两个相同大小区域时，一个区域的值总是大于另一区域的值，推出

then we divide all elements in $\{\text{A},\text{B}\}$ into two parts with equal length, and one part is always greater than the other. Then

\text{median} = \frac{\text{max}(\text{left}\_\text{part}) + \text{min}(\text{right}\_\text{part})}{2}median=​2​​max(left_part)+min(right_part)​​

满足这两个条件，只需要满足这两个条件即可：

To ensure these two conditions, we just need to ensure:

1. $i+j=m-i+n-j$ (or: $m-i+n-j+1$)
   if $n\ge m$, we just need to set:  i=0∼m, j=m+n+12−i i=0∼m, j=m+n+12−i

2. $\text{B}[j-1]\le \text{A}[i]$ and $\text{A}[i-1]\le \text{B}[j]$

学习了

right_A