127 lines
4.1 KiB
Java
127 lines
4.1 KiB
Java
/*
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* The MIT License (MIT)
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*
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* Copyright © 2020 xrv <xrg@live.com>
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*
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* Permission is hereby granted, free of charge, to any person obtaining a copy
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* of this software and associated documentation files (the "Software"), to deal
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* in the Software without restriction, including without limitation the rights
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* to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
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* copies of the Software, and to permit persons to whom the Software is
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* furnished to do so, subject to the following conditions:
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*
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* The above copyright notice and this permission notice shall be included in
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* all copies or substantial portions of the Software.
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*
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* THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
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* IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
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* FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
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* AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
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* LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
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* OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
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* THE SOFTWARE.
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*/
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package io.github.ehlxr.leetcode;
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/**
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* 4. 寻找两个正序数组的中位数
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*
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* @author ehlxr
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* @since 2020-09-27 21:12.
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*/
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public class Solution4 {
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/**
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* 解法一
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* 简单粗暴，先将两个数组合并，两个有序数组的合并也是归并排序中的一部分。然后根据奇数，还是偶数，返回中位数。
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*/
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public static double findMedianSortedArrays(int[] nums1, int[] nums2) {
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int[] result = new int[nums1.length + nums2.length];
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int x = 0, y = 0, z = 0;
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while (x < nums1.length || y < nums2.length) {
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if (x >= nums1.length) {
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result[z] = nums2[y];
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y++;
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z++;
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continue;
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}
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if (y >= nums2.length) {
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result[z] = nums1[x];
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x++;
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z++;
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continue;
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}
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if (nums1[x] < nums2[y]) {
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result[z] = nums1[x];
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x++;
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} else {
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result[z] = nums2[y];
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y++;
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}
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z++;
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}
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int m = result.length / 2;
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if (result.length % 2 == 0) {
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return (result[m] + result[m - 1]) / 2.0;
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} else {
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return result[m];
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}
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}
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/**
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* 用 len 表示合并后数组的长度，
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* 如果是奇数，我们需要知道第 （len+1）/2 个数就可以了，如果遍历的话需要遍历 int(len/2 ) + 1 次。
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* 如果是偶数，我们需要知道第 len/2 和 len/2+1 个数，也是需要遍历 len/2+1 次。
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*
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* 所以遍历的话，奇数和偶数都是 len/2+1 次。
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*
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* https://leetcode-cn.com/problems/median-of-two-sorted-arrays/solution/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by-w-2/
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*/
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public static double findMedianSortedArrays2(int[] nums1, int[] nums2) {
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int l1 = nums1.length, l2 = nums2.length,
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left = 0, right = 0,
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len = l1 + l2,
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x = 0, y = 0;
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for (int i = 0; i < len/2 +1; i++) {
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left = right;
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if (x >= nums1.length) {
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right = nums2[y];
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y++;
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continue;
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}
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if (y >= nums2.length) {
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right = nums1[x];
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x++;
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continue;
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}
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if (nums1[x] < nums2[y]) {
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right = nums1[x];
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x++;
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} else {
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right = nums2[y];
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y++;
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}
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}
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if (len % 2 == 0) {
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return (left + right) / 2.0;
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} else {
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return right;
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}
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}
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public static void main(String[] args) {
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System.out.println(findMedianSortedArrays(new int[]{}, new int[]{2, 3}));
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System.out.println(findMedianSortedArrays2(new int[]{1, 4}, new int[]{2, 3}));
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System.out.println(findMedianSortedArrays2(new int[]{3}, new int[]{1, 4, 6}));
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}
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}
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