88. Merge Sorted Array

问题描述

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.

题目链接：

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思路分析

给出两个已经排序的数组，将数组nums1变成两个数组合并后的已排序数组。可以认为nums1在初始化的开始就有足够的空间容纳所有的m+n个元素

代码

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int i = m - 1, j = n - 1;
        while(i >= 0 && j >= 0){//倒序循环，从大到小排序
            if (nums1[i] >= nums2[j]){
                nums1[i+j+1] = nums1[i];
                i--;
            }
            else{
                nums1[i+j+1] = nums2[j];
                j--;
            }   
        }
        for(int i = 0; i <= j; i++){//nums1为空或者nums2还有剩下的元素比nums1[0]还小，就放在最前面
                nums1[i] = nums2[i];
        }
    }
};

时间复杂度：O(m+n)

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反思

虽然很丑陋，但是蜜汁快，3ms过。