leetcode-4-OK

address

double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size){
	int index1,index2;
	if( (nums1Size+nums2Size)%2 ==1){
		index1 = (nums1Size+nums2Size+1)/2;
		index2 = index1;
	}else{
		index1 = (nums1Size+nums2Size)/2;
		index2 = index1+1;
	}
	double an1,an2;
	int for1=0,for2=0,count=0,num;

	while(count != index1){
		if((for1<nums1Size) && (for2<nums2Size)){	// 情况1
			if(nums1[for1]<nums2[for2]){
				count++;
				num = nums1[for1++];
			}else{
				count++;
				num = nums2[for2++];
			}
		}else if(for1==nums1Size){	// 情况2
			count++;
			num = nums2[for2++];
		}else if(for2==nums2Size){	// 情况3
			count++;
			num = nums1[for1++];
		}
	}
	an1 = num;
	if(index1 == index2)
		return an1;			// 奇数总数
	if((for1<nums1Size) && (for2<nums2Size)){	// 情况1
		if(nums1[for1]<nums2[for2]){
			count++;
			an2 = nums1[for1++];
		}else{
			count++;
			an2 = nums2[for2++];
		}
	}else if(for1==nums1Size){
		count++;
		an2 = nums2[for2++];
	}else if(for2==nums2Size){
		count++;
		an2 = nums1[for1++];
	}
	
	return (an1+an2)/2;
}

leetcode-4-OK

原文：https://www.cnblogs.com/gallien/p/14290908.html