LeetCode：4. Median of Two Sorted Arrays 两个排序数组的中值

最新推荐文章于 2019-09-21 11:46:00 发布

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本文介绍了一种在O(log(m+n))的时间复杂度内找到两个已排序数组中位数的方法。通过两种不同的实现方式，展示了如何合并并排序这两个数组来找到中位数，同时也提供了一种更高效的解决方案，避免了完整的数组合并。

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一、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)).

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
二、代码

简单合并，排序，取中值

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {	
        int *it;
        int m = nums1.size()?nums1.size():0;
        int n = nums2.size()?nums2.size():0;
        int *merge = new int[m+n];

        if(!nums1.empty()){
            it = &*nums1.begin();
            memcpy(merge,it,sizeof(int)*m);   
        }
        if(!nums2.empty()){
            it = &*nums2.begin();
            memcpy(merge+m,it,sizeof(int)*n);  
        }
        sort(merge,merge+m+n);
        //注意数组下标和计算的下标区别
        return (m+n)%2?1.0*merge[(m+n)>>1]:(merge[(m+n-1)>>1]+merge[(m+n)>>1])/2.0;
        
    }
};

class Solution {
    public double findMedianSortedArrays(int[] A, int[] B) {
        int m = A.length;
        int n = B.length;
        if (m > n) {
            int[] temp = A; A = B; B = temp;
            int tmp = m; m = n; n = tmp;
        }
        
        int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
        while (iMin <= iMax) {
            int i = (iMin + iMax) / 2;
            int j = halfLen - i;
            if (i < iMax && B[j-1] > A[i]){
                iMin = i + 1; // i is too small
            }else if (i > iMin && A[i-1] > B[j]) {
                iMax = i - 1; // i is too big
            }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;
    }
}