leetcode Median of Two Sorted Arrays(Java)

题目链接：点击打开链接

类型：合并数组

解法：遍历数组，排列生成新的数组

public class Solution {    public double findMedianSortedArrays(int[] nums1, int[] nums2) {        int len1 = nums1.length;        int len2 = nums2.length;        int len3 = len1 + len2;        int[] nums3 = new int[len3];                for (int i=0, j=0, k=0 ; k<len3 ;)        {            if (i < len1 && j < len2)            {            nums3[k] = (nums1[i] < nums2[j]) ? nums1[i++] : nums2[j++];            ++k;            }            if (i == len1 && j < len2 && i < len3)            {            nums3[k] = nums2[j++];            ++k;            }            if (i < len1 && j == len2 && j < len3)            {            nums3[k] = nums1[i++];            ++k;            }        }                double result = (len3%2==0)?((nums3[len3/2]+nums3[len3/2 - 1])*1.0/2):nums3[len3/2];                return result;    }}

此外，在搜索题解时，找到了一种通过寻找第k小数进行求解的方法，现记录如下：

public class Solution {    public double findKelemnt(int[] a, int astart, int aend, int[] b, int bstart, int bend, int k){int m = aend - astart + 1;int n = bend - bstart + 1;if (m > n)return findKelemnt(b, bstart, bend, a, astart, aend, k);if (m == 0)return b[k - 1];if (k == 1)return Math.min(a[astart], b[bstart]);int partA = Math.min(k / 2, m);int partB = k - partA;if (a[astart + partA - 1] < b[bstart + partB - 1]){return findKelemnt(a, astart + partA, aend, b, bstart, bend, k - partA);} else if (a[astart + partA - 1] > b[bstart + partB - 1]){return findKelemnt(a, astart, aend, b, bstart + partB, bend, k - partB);} else{return a[astart + partA - 1];}}    public double findMedianSortedArrays(int[] nums1, int[] nums2) {double retDouble = 0.0;int len1 = nums1.length;int len2 = nums2.length;int len3 = len1 + len2;int k = len3 / 2;if (len3 % 2 == 0){retDouble = (findKelemnt(nums1, 0, len1 - 1, nums2, 0, len2 - 1, k + 1)+ findKelemnt(nums1, 0, len1 - 1, nums2, 0, len2 - 1, k)) / 2;} else{retDouble = findKelemnt(nums1, 0, len1 - 1, nums2, 0, len2 - 1, k + 1);}return retDouble;    }}