LeetCode 4. Median of Two Sorted Arrays

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本文解析了LeetCode第4题“寻找两个有序数组的中位数”的算法实现，介绍了两种不同的解决方案，并对代码进行了详细解释。该问题要求在O(log(m+n))的时间复杂度内找到两个有序数组的中位数。

LeetCode 4. Median of Two Sorted Arrays

Description

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

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Code

java

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        double ans = 0.0;
        int len1 = nums1.length, len2 = nums2.length;
        int len = len1 + len2;
        int mid = len / 2, cnt = -1, temp = 0, pre = 0;
        int i = 0, j =0;
        while(i < len1 || j < len2) {
            if((j == len2) || (i < len1 && nums1[i] < nums2[j])) {
                temp = nums1[i++];
            } else {
                temp = nums2[j++];
            }
            cnt++;
            if(cnt == mid - 1) {
                pre = temp;
            } else if(cnt == mid) {
                ans = temp;
                break;
            }
        }
        if(len % 2 == 0) {
            ans = (ans + pre) / 2;
        }
        return ans;
    }
}

Official Solution

class Solution {
    public double findMedianSortedArrays(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;
        }
        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;
    }
}