LeetCode--Median of Two Sorted Arrays

#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;
    }
}