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; } }