面试经典150题——Day1

一、题目

88.Merge Sorted Array

You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.

Merge nums1 and nums2 into a single array sorted in non-decreasing order.

The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].
Example 3:

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.

Constraints:

nums1.length == m + n
nums2.length == n
0 <= m, n <= 200
1 <= m + n <= 200
-109 <= nums1[i], nums2[j] <= 109

Follow up: Can you come up with an algorithm that runs in O(m + n) time?

来源：leetcode

二、我的笨方法

class Solution {
public:
    int min(int a,int b){
        if(a < b) return a;
        else return b;
    }
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        if(n == 0) return;
        vector<int> res;
        int i = 0,j = 0;
        for(int k = 0;k < (m + n);k++){
            if(i < m && j < n){
                res.push_back(min(nums1[i],nums2[j]));
                if(res[k] == nums1[i]) i++;
                else j++;
            }
            else if(i == m && j < n){
                res.push_back(nums2[j]);
                j++;
            }
            else if(i < m && j == n){
                res.push_back(nums1[i]);
                i++;
            }
        }
        for(int i = 0;i < (m + n);i++){
            nums1[i] = res[i];
        }
        return;
    }
};
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三、更好的方法

直接利用sort函数

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        for(int i = m,j = 0;i < m + n;i++,j++){
            nums1[i] = nums2[j];
        }
        sort(nums1.begin(),nums1.end());
    }
};
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