给定两个有序数组arr1和arr2，两个数组长度都为N，求两个数组中所有数的上中位数。 例如： arr1 = {1,2,3,4}; arr2 = {3,4,5,6}; 一共8个数则上中位数是第4个数，所以返回3。 arr1 = {0,1,2}; arr2 = {3,4,5}; 一共6个数则上中位数是第3个数，所以返回2。 要求：时间复杂度O(logN) public static int getUpMedian(int[] arr1, int[] arr2) { if (arr1 == null || arr2 == null || arr1.length != arr2.length) { throw new RuntimeException("Your arr is invalid!"); } return findProcess(arr1, 0, arr1.length - 1, arr2, 0, arr2.length - 1); } public static int findProcess(int[] arr1, int start1, int end1, int[] arr2, int start2, int end2) { if (start1 == end1) { return Math.min(arr1[start1], arr2[start2]); } // 元素个数为奇数，则offset为0；元素个数为偶数，则offset为1； int offset = ((end1 - start1 + 1) & 1) ^ 1; int mid1 = (start1 + end1) / 2; int mid2 = (start2 + end2) / 2; if (arr1[mid1] > arr2[mid2]) { return findProcess(arr1, start1, mid1, arr2, mid2 + offset, end2); } else if (arr1[mid1] < arr2[mid2]) { return findProcess(arr1, mid1 + offset, end1, arr2, start2, mid2); } else { return arr1[mid1]; } }