Leetcode 4 Median of Two Sorted Arrays

最新推荐文章于 2023-06-12 16:30:30 发布

原创最新推荐文章于 2023-06-12 16:30:30 发布 · 209 阅读

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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有序数组中位数的方法。通过调整两个数组的指针位置，比较并合并元素，最终找到中位数。算法分为直接合并和优化后的二分查找两种实现方式。

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链接：https://leetcode-cn.com/problems/median-of-two-sorted-arrays/solution/

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 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

class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int len1 = nums1.length;
        int len2 = nums2.length;
        int length = len1 + len2;
        int flag = 0;
        int[] nums = new int[length]; 
        int n = 0,m = 0;
        int i = 0;
        int j = length / 2;
        for(i = 0 ; i < length;i++){
            if(m < len2 && n < len1 && nums1[n] > nums2[m]){
                nums[i] = nums2[m];
                m++;
            }else if(n < len1 && m < len2 && nums1[n] <= nums2[m]){
                nums[i] = nums1[n];
                n++;
            }else if(n == len1 && m != len2){
                nums[i] = nums2[m];
                m++;
            }else if(m == len2 && n != len1){
                nums[i] = nums1[n];
                n++;
            }
        }
        if(length % 2 == 0){
            return (nums[j] + nums[j - 1]) / 2.0;
        }else{
            return nums[j];
        }
    }
}

时间复杂度为：O（n + m）

空间复杂度为：O（n）

官方题解：

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;
    }
}