【LeetCode】88 Merge Sorted Array

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本文介绍了一种快速合并两个已排序数组成一个有序数组的方法。通过逆向比较元素，算法实现了一次性合并，避免了多次遍历。详细阐述了两种方法，包括其时间复杂度和空间复杂度分析，旨在提升对数据合并的理解和实现效率。

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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.

Summary:
1. My mind is narrow. I always want to realize the comparison from the small number to the big number. Why i can’t get the idea that reverse the order of comparison.
2. I have not a clarify logic to write the solution for the problem. Just like the Method 2 , it takes my 2 or 3 hours.

*Method 1:(Runtime: 0 ms)

public class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int index = m+n-1;
        int index1 = m - 1;
        int index2 = n - 1;

        while(index1 >= 0 && index2 >= 0){
            if(nums1[index1] >= nums2[index2]){
                nums1[index--] = nums1[index1--];
            }else{
                nums1[index--] = nums2[index2--];  
            }
        }

        while(index1 >= 0){
            nums1[index--] = nums1[index1--];
        }

        while(index2 >= 0){
            nums1[index--] = nums2[index2--];
        }
    }
}

Method 2:(Runtime : 4ms)

public class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        for (int j = 0; j < n; j++) {
            for (int i = 0; i < m+1; i++) {
                if (nums2[j] <= nums1[i]) {
                    System.arraycopy(nums1, i, nums1, i + 1, m - i);
                    m++;
                    nums1[i] = nums2[j];
                    break;
                }else if(i>=m){
                    nums1[i] = nums2[j];
                    m++;
                    break;
                }
            }
        }
    }
}