2个有序数组，寻找他们合并之后的中位数_两个数据库查找中位数不合并-优快云博客

本文介绍了一种方法，用于在两个已排序的整数数组中找到中间元素，即使数组长度不相等。该算法通过比较中间元素并进行适当的移动来优化查找过程。

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public class MiddleGetter {

	public static int getMiddle(int[] sortA, int[] sortB) {
		if (sortA == null && sortB == null) {
			throw new NullPointerException();
		} else if (sortA == null || sortB == null) {
			return (sortA == null) ? sortA[sortA.length / 2] : sortB[sortB.length / 2];
		}

		int middleInA = sortA.length / 2;
		int middleInB = sortB.length / 2;
		int lengthA = sortA.length;
		int lengthB = sortB.length;

		if (sortA[middleInA] == sortB[middleInB]) {
			return sortA[middleInA];
		} else {
			int maxMiddle = sortA[middleInA] > sortB[middleInB] ? 0 : 1;
			int lastMove = maxMiddle;

			if (maxMiddle == 0) {
				--middleInA;
			} else {
				--middleInB;
			}
			do {
				if (lastMove == maxMiddle) {
					if (lastMove == 0) {
						if (middleInA == -1) {
							return sortB[middleInB];
						}
						if (sortA[middleInA] <= sortB[middleInB]) {
							return sortB[middleInB];
						} else {
							middleInB++;
							lastMove = 1;
							continue;
						}
					}
					if (lastMove == 1) {
						if (middleInB == -1) {
							return sortA[middleInA];
						}
						if (sortB[middleInB] <= sortA[middleInA]) {
							return sortA[middleInA];
						} else {
							middleInA++;
							lastMove = 0;
							continue;
						}
					}
				} else {
					if (lastMove == 0) {
						if (middleInA == lengthA) {
							return sortB[middleInB];
						}
						if (sortA[middleInA] >= sortB[middleInB]) {
							return sortB[middleInB];
						} else {
							middleInB--;
							lastMove = 1;
							continue;
						}
					}
					if (lastMove == 1) {
						if (middleInB == lengthB) {
							return sortA[middleInA];
						}
						if (sortB[middleInB] >= sortA[middleInA]) {
							return sortA[middleInA];
						} else {
							middleInA--;
							lastMove = 0;
							continue;
						}
					}
				}
			} while (true);
		}
	}

	public static void main(String[] args) {
		int[] A = { 9 };
		int[] B = { 0, 1 };
		System.out.println(getMiddle(A, B));

	}

}