【LeetCode】88 Merge Sorted Array

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;                }            }        }    }}