leetcode-DAY2-[167,189.283]

总结今天所用到的vector的知识

insert
[ ] iterator insert (iterator position, const value_type& val);在position位置插入val，返回迭代器；
[ ] void insert (iterator position, size_type n, const value_type& val);在position位置插入n个val，没有返回值；
[ ] void insert (iterator position, InputIterator first, InputIterator last);在position位置插入从fisrt迭代器开始到last迭代器位置的元素，包括fisrt，不包括last。

erase
[ ] iterator erase (iterator position);删除position位置的元素，随机返回迭代器，小心野指针；
[ ] iterator erase (iterator first, iterator last);删除从first到last位置的元素，包括first，不包括last。
push_backpush最后一个元素
pop_backpop最后一个元素
remove
1. Move Zeroes

Given an array nums, write a function to move all 0’s to the end of it while maintaining the relative order of the non-zero elements.

Example:

Input: [0,1,0,3,12]
Output: [1,3,12,0,0]
Note:

You must do this in-place without making a copy of the array.
Minimize the total number of operations.

void Solution283::moveZeroes(vector<int>& nums)
{
    vector<int>::iterator pre = nums.begin();
    vector<int>::iterator i;
    int k = 0;
    while (pre < nums.end())
    {
        if (*pre == 0)
        {
            ++k;
            i = pre;
            nums.erase(i);
        }
        else
            pre++;
    }
    nums.insert(nums.end(), k, 0);
}

收获：
1. while循环每次判断的时候，应该判断nums.end()因为如果提前赋值it=nums.end()-1，每次删除数组中元素，会导致it迭代器越界
2. 在循环中执行nums.erase(it)时，执行完毕会随机返回迭代器，如果提前赋值会使迭代器称为野指针，便不能执行++操作
3. insert在末尾插入位置为nums.end()

Rotate Array

Given an array, rotate the array to the right by k steps, where k is non-negative.

Example 1:

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Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]


void Solution189::rotate(vector<int>& nums, int k)
{
    if (nums.size() < k) k = k % nums.size();
    nums.insert(nums.begin(), nums.end() - k, nums.end());
    nums.erase(nums.end() - k, nums.end());
}

收获：
1. 这里学习了别人的代码，在数组长度小于k的时候重新对k取余赋值，实际上前面的数组长度的整数倍的转换都是多余的
2. 这时候就可以利用insert方法，先插入，再统一删除

Two Sum II - Input array is sorted

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.

Note:

Your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.
Example:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

    vector<int>::iterator p = numbers.begin();
    vector<int>::iterator q = numbers.end();
    --q;
    vector<int> index;
    while (p < q)
    {
        if ((*p + *q) < target)
        {
            ++p;
        }
        else if ((*p + *q) > target)
        {
            --q;
        }
        else
        {
            index.push_back(p - numbers.begin() + 1);
            index.push_back(q - numbers.begin() + 1);
            break;
        }
    }
    return index;

收获：
1. 算法思想：两个指针指向前后，分别判断和target的大小，如果大后面指针前移，如果小前面指针后移

leetcode-DAY2-[167,189.283]

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