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#Median of Two Sorted Arrays
##题目
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)).
You may assume nums1 and nums2 cannot be both empty.
###Example
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
###Example
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
##分析
这题要求时间复杂度为O(log (m+n)).所以我们需要在遍历一次的一半时间内需要将结果找出来。而且我们不需要额外的申请空间,我们采取的策略是两个数组同时遍历,遍历的过程中同时比较元素的大小,这样我们就可以找出前O(log (m+n))个元素,然后就是我们需要寻找的结果。
##源码
double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
int totalSize = nums1Size + nums2Size;
int medianSize = totalSize/2;
int a1 = 0, a2 = 0;
double result = 0;
int t = 1;
//当第一个数组为空的时候,再分别讨论奇偶
if(nums1Size == 0) {
if(totalSize%2 == 0) {
int tag1,tag2;
for(int i = 0; i <= medianSize; i++) {
if(i == medianSize -1) {
tag1 = nums2[i];
}
if(i == medianSize) {
tag2 = nums2[i];
}
}
return (double)(tag1+tag2)/2;
} else {
int tag1,tag2;
for(int i = 0; i <= medianSize; i++) {
if(i == medianSize) {
tag1 = nums2[i];
}
}
return tag1;
}
}
//当第二个数组为空的时候,再分别讨论奇偶
if(nums2Size == 0) {
if(totalSize%2 == 0) {
int tag1,tag2;
for(int i = 0; i <= medianSize; i++) {
if(i == medianSize -1) {
tag1 = nums1[i];
}
if(i == medianSize) {
tag2 = nums1[i];
}
}
return (double)(tag1+tag2)/2;
} else {
int tag1,tag2;
for(int i = 0; i <= medianSize; i++) {
if(i == medianSize) {
tag1 = nums1[i];
}
}
return tag1;
}
}
if(totalSize%2 == 0) {
int tag1 = 0, tag2 = 0;
for(int i = 0; i <= medianSize; i++) {
if((nums1[a1] <= nums2[a2])&&(a1 < nums1Size)||a2 == nums2Size) {
if(i == medianSize - 1) {
tag1 = nums1[a1];
}
if(i == medianSize) {
tag2 = nums1[a1];
}
a1++;
} else{
if(i == medianSize - 1) {
tag1 = nums2[a2];
}
if(i == medianSize) {
tag2 = nums2[a2];
}
a2++;
}
}
//printf("%d\n", tag1);
return (double)(tag1+tag2)/2;
} else {
int tag1 = 0;
for(int i = 0; i <= medianSize; i++){
if((nums1[a1] <= nums2[a2])&&(a1 < nums1Size)||a2 == nums2Size) {
if(i == medianSize ) {
tag1 = nums1[a1];
}
a1++;
} else {
if(i == medianSize) {
tag1 = nums2[a2];
}
a2++;
}
}
return tag1;
}
}