给定两个有序数组arr1和arr2，在给定一个整数k，返回两个数组的所有数中第K小的数。 例如： arr1 = {1,2,3,4,5}; arr2 = {3,4,5}; K = 1; 因为1为所有数中最小的，所以返回1； arr1 = {1,2,3}; arr2 = {3,4,5,6}; K = 4; 因为3为所有数中第4小的数，所以返回3； 要求：如果arr1的长度为N，arr2的长度为M，时间复杂度请达到O(log(min{M,N}))。 public static int findKthNum(int[] arr1, int[] arr2, int kth) { if (arr1 == null || arr2 == null) { throw new RuntimeException("Your arr is invalid!"); } if (kth < 1 || kth > arr1.length + arr2.length) { throw new RuntimeException("K is invalid!"); } int[] longArr = arr1.length >= arr2.length ? arr1 : arr2; int[] shortArr = arr1.length < arr2.length ? arr1 : arr2; int lenL = longArr.length; int lenS = shortArr.length; if (kth <= lenS) { return getUpMedian(shortArr, 0, kth - 1, longArr, 0, kth - 1); } if (kth > lenL) { if (shortArr[kth - lenL - 1] >= longArr[lenL - 1]) { return shortArr[kth - lenL - 1]; } if (longArr[kth - lenS - 1] >= shortArr[lenS - 1]) { return longArr[kth - lenS - 1]; } return getUpMedian(shortArr, kth - lenL, lenS - 1, longArr, kth - lenS, lenL - 1); } if (longArr[kth - lenS - 1] >= shortArr[lenS - 1]) { return longArr[kth - lenS - 1]; } return getUpMedian(shortArr, 0, lenS - 1, longArr, kth - lenS, kth - 1); } public static int getUpMedian(int[] arr1, int start1, int end1, int[] arr2, int start2, int end2) { if (start1 == end1) { return Math.min(arr1[start1], arr2[start2]); } int offset = ((end1 - start1 + 1) & 1) ^ 1; int mid1 = (start1 + end1) / 2; int mid2 = (start2 + end2) / 2; if (arr1[mid1] > arr2[mid2]) { return getUpMedian(arr1, start1, mid1, arr2, mid2 + offset, end2); } else if (arr1[mid1] < arr2[mid2]) { return getUpMedian(arr1, mid1 + offset, end1, arr2, start2, mid2); } else { return arr1[mid1]; } }