[LeetCode] 4. 寻找两个正序数组的中位数

方法一:

class Solution {
    public double findMedianSortedArrays(int[] A, int[] B) {
    int m = A.length;
    int n = B.length;
    int len = m + n;
    int left = -1, right = -1;
    int aStart = 0, bStart = 0;
    for (int i = 0; i <= len / 2; i++) {
        left = right;
        if (aStart < m && (bStart >= n || A[aStart] < B[bStart])) {
            right = A[aStart++];
        } else {
            right = B[bStart++];
        }
    }
    if ((len & 1) == 0)
        return (left + right) / 2.0;
    else
        return right;
}
}

============================

 方法二:

每次都去掉数组中的k/2的数量值,直到最后一个,复杂度只有log

public double findMedianSortedArrays(int[] nums1, int[] nums2) {
    int n = nums1.length;
    int m = nums2.length;
    int left = (n + m + 1) / 2;
    int right = (n + m + 2) / 2;
    //将偶数和奇数的情况合并,如果是奇数,会求两次同样的 k 。
    return (getKth(nums1, 0, n - 1, nums2, 0, m - 1, left) + getKth(nums1, 0, n - 1, nums2, 0, m - 1, right)) * 0.5;  
}
    
    private int getKth(int[] nums1, int start1, int end1, int[] nums2, int start2, int end2, int k) {
        int len1 = end1 - start1 + 1;
        int len2 = end2 - start2 + 1;
        //让 len1 的长度小于 len2,这样就能保证如果有数组空了,一定是 len1 
        if (len1 > len2) return getKth(nums2, start2, end2, nums1, start1, end1, k);
        if (len1 == 0) return nums2[start2 + k - 1];

        if (k == 1) return Math.min(nums1[start1], nums2[start2]);

        int i = start1 + Math.min(len1, k / 2) - 1;
        int j = start2 + Math.min(len2, k / 2) - 1;

        if (nums1[i] > nums2[j]) {
            return getKth(nums1, start1, end1, nums2, j + 1, end2, k - (j - start2 + 1));
        }
        else {
            return getKth(nums1, i + 1, end1, nums2, start2, end2, k - (i - start1 + 1));
        }
    }

作者:windliang
链接:https://leetcode-cn.com/problems/median-of-two-sorted-arrays/solution/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by-w-2/
来源:力扣(LeetCode)
著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。

牛啊!

 

posted @ 2020-05-25 00:18  doyi  阅读(629)  评论(0编辑  收藏  举报