数据结构和算法笔记2：二分法

二分法网上有两种写法，一种左闭右闭，一种左闭右开，个人习惯左闭右闭的写法，

有序数组查找数

这是标准二分法，对应力扣的704. 二分查找：

• 求值为target的索引

int search(vector<int>& nums, int target) {
    int left = 0; 
    int right = nums.size();
    while (left < right)
    {
        int mid = left + (right - left) / 2;
        if (nums[mid] > target)
            right = mid;
        else if (nums[mid] < target)
            left = mid + 1;
        else
            return mid;
    }
    return -1;
}

求左右边界

二分法还可以求第一个大于target的索引和第一个小于target的索引

• 求第一个大于target的值的索引（右边界）

int getRightIndex(vector<int>& nums, int target)
{
    int left = 0; 
    int right = nums.size() - 1;
    int rightBorder = -2;
    while (left <= right)
    {
        int mid = left + (right - left) / 2;
        if (nums[mid] > target)
            right = mid - 1;
        else
        {
            left = mid + 1;
            rightBorder = left;
        }
    }
    return rightBorder;
}

力扣的35. 搜索插入位置也可以用这个思路解决：

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        int left = 0; 
        int right = nums.size() - 1;
        int rightBorder = 0;
        while (left <= right)
        {
            int mid = left + (right - left) / 2;
            if (nums[mid] > target)
                right = mid - 1;
            else
            {
                left = mid + 1;
                rightBorder = left;
            }
        }
        if (rightBorder == 0)
        {
            return 0;
        }
        else
        {
            if (nums[rightBorder - 1] != target)
                return rightBorder;
            else
                return rightBorder - 1;
                
        }

    }
};

当然只有一个target也有更简洁的写法，左闭右闭的写法里最后的left就是右边界了：

class Solution {
public:
    int searchInsert(vector<int>& nums, int target) {
        int left = 0;
        int right = nums.size() - 1;
        while (left <= right)
        {
            int mid = left + (right - left) / 2;
            if (nums[mid] > target)
                right = mid - 1;
            else if (nums[mid] < target)
                left = mid + 1;
            else
                return mid;
        }
        return left;
    }
};

• 求第一个小于target的值的索引（左边界）

int getLeftIndex(vector<int>& nums, int target)
{
    int left = 0; 
    int right = nums.size() - 1;
    int leftBorder = -2;
    while (left <= right)
    {
        int mid = left + (right - left) / 2;
        if (nums[mid] < target)
            left = mid + 1;
        else
        {
            right = mid - 1;
            leftBorder = right;
        }
    }
    return leftBorder;
}

使用上面两种方法求第一个大于target的索引和第一个小于target的索引对应的是力扣的34. 在排序数组中查找元素的第一个和最后一个位置。第一个target出现的索引和最后一个target出现的索引，对应的就是左边界+1和右边界-1，题目的完整代码：

class Solution {
public:
    int getRightIndex(vector<int>& nums, int target)
    {
        int left = 0; 
        int right = nums.size() - 1;
        int rightBorder = -2;
        while (left <= right)
        {
            int mid = left + (right - left) / 2;
            if (nums[mid] > target)
                right = mid - 1;
            else
            {
                left = mid + 1;
                rightBorder = left;
            }
        }
        return rightBorder;
    }

    int getLeftIndex(vector<int>& nums, int target)
    {
        int left = 0; 
        int right = nums.size() - 1;
        int leftBorder = -2;
        while (left <= right)
        {
            int mid = left + (right - left) / 2;
            if (nums[mid] < target)
                left = mid + 1;
            else
            {
                right = mid - 1;
                leftBorder = right;
            }
        }
        return leftBorder;
    }

    vector<int> searchRange(vector<int>& nums, int target) {
        int leftBorder = getLeftIndex(nums, target);
        int rightBorder = getRightIndex(nums, target);
        if (leftBorder == -2 || rightBorder == -2)
            return {-1, -1};
        if (rightBorder - leftBorder > 1)
            return {leftBorder + 1, rightBorder - 1};
        return {-1, -1};

        
    }
};