1、問題

Given two sorted integer arrays?nums1?and?nums2, merge?nums2?into?nums1?as one sorted array.

Note:
You may assume that?nums1?has enough space (size that is greater or equal to?m?+?n) to hold additional elements from?nums2. The number of elements initialized in?nums1?and?nums2?are?m?and?n?respectively.

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2、代碼實(shí)現(xiàn)

package leetcode.chenyu.test;public class MergeSortedArray {public static void main(String[] args) {int a[] = new int[10];a[0] = -1;a[1] = 0;a[2] = 4;a[3] = 7;int b[] = {2, 5};int len = a.length;for (int x : a) System.out.print(x);merge(a, 10, b, 2);for (int x : a) System.out.print(x);}public static void merge(int[] nums1, int m, int[] nums2, int n) {if (nums1 == null || m == 0 || n == 0) return;int i = 0, j = 0, k = 0;int index = 0;int nums3[] = new int[m];for (int x = m - 1; x>= 0; x--) {if (x - 1 > 0) if (nums1[x] == 0 && nums1[x - 1] != 0) {index = x;break;}}int value = nums1[index];System.out.println("index is:" + index);while (j < n && i < index) {if (nums1[i] <= nums2[j]) {System.out.print("if i is " + i + "k is " + k + "nums1[" + i + "]" + nums1[i] + "\n");nums3[k] = nums1[i];i++;} else {System.out.print("if j is " + j + "k is " + k + "nums2["+ j + "]" + nums2[j] + "\n");nums3[k] = nums2[j];j++;}k++;}System.out.println("i + 1 < m" + (i + 1 < m));System.out.println("i is :" + i);System.out.println("nums1[i]" + nums1[i]);System.out.println("nums1[index]" + nums1[index]);System.out.println("nums1[i + 1] != nums1[index]" + (nums1[i + 1] != nums1[index]));if (i + 1 < m && nums1[i] != nums1[index]) {System.out.println("if if");for (int f = i; f < index; f++) {nums3[k] = nums1[f];}}for (int x : nums3) {System.out.print(x);}for (int h = 0; h < nums3.length; h++) {nums1[h] = nums3[h];} // nums1 = nums3;} }
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3、結(jié)果

-1047000000index is:4 if i is 0k is 0nums1[0]-1 if i is 1k is 1nums1[1]0 if j is 0k is 2nums2[0]2 if i is 2k is 3nums1[2]4 if j is 1k is 4nums2[1]5 i + 1 < mtrue i is :3 nums1[i]7 nums1[index]0 nums1[i + 1] != nums1[index]false if if -1024570000-1024570000
? 提到上去會(huì)有下標(biāo)越界異常，好吧，后面再來(lái)分析為什么出錯(cuò)，今天先記錄到這里

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