Leetcode题解 88. Merge Sorted Array（todo）

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.

投机取巧的做法：把nums2放再nums1后面。对nums1用Java的函数来排序即可。

public class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
for(int i=m;i<m+n;i++){
nums1[i]=nums2[i-m];
}
Arrays.sort(nums1);
}
}

新的解法等复习完归并排序再来搞。

新的解法来了。。。

两个数组均从后往前遍历即可。

public class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int pos=m+n-1;
int i=m-1;
int j=n-1;
while(j>=0){
if(i>=0&&nums1[i]>nums2[j]){
nums1[pos--]=nums1[i--];
}else{
nums1[pos--]=nums2[j--];
}
}
}
}