【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）
74. Search a 2D Matrix
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
-
Integers in each row are sorted from left to right.
-
The first integer of each row is greater than the last integer of the previous row.
Example 1:
Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 3
Output: true
Example 2:
Input:
matrix = [
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
target = 13
Output: false
/*
问题：在有序矩阵中查找某数（每行增序，且行首元素大于前一行的行尾元素，蛇形有序）
方法：将二维数组看成一维数组用二分查找即可
*/
class Solution
{
public:
bool searchMatrix(vector<vector<int>>& matrix, int target)
{
if(matrix.empty() || matrix[0].empty()) return false;
int rows = matrix.size();
int cols = matrix[0].size();
int left = 0, right = rows*cols - 1;
while(left <= right)
{
int mid = (left+right) / 2;
if(matrix[mid/cols][mid%cols] < target) left = mid+1;
else if(matrix[mid/cols][mid%cols] > target) right = mid - 1;
else return true;
}
return false;
}
};
240. Search a 2D Matrix II
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:
-
Integers in each row are sorted in ascending from left to right.
-
Integers in each column are sorted in ascending from top to bottom.
Consider the following matrix:
[
[1, 4, 7, 11, 15],
[2, 5, 8, 12, 19],
[3, 6, 9, 16, 22],
[10, 13, 14, 17, 24],
[18, 21, 23, 26, 30]
]
Example 1:
Input: matrix, target = 5Output: true
Example 2:
Input: matrix, target = 20
Output: false
//问题：有序矩阵中查找(非蛇形有序)
/*
方法一：扫描行，每行进行二分查找
O(r*logc)
*/
class Solution
{
public:
bool searchMatrix(vector<vector<int>>& a, int target)
{
if(a.empty() || a[0].empty()) return false; //若为空，则返回false
int row = a.size(); //行数
int col = a[0].size(); //列数
for(int i = 0; i < row; i++)//遍历行，target要处于行首和行尾元素之间
{
if(a[i][0]<=target&&a[i][col-1] >= target) //不能放在for语句里，因为一旦条件不满足就跳出整个for循环了
{
if(a[i][0] == target || a[i][col-1] == target) return true;//如果行首或行尾元素等于target则返回true
// cout<<"行数为"<<i<<endl;
int left = 0, right = col-1; //遍历列，用二分查找算法查找
while(left<=right)
{
int mid = (left+right)/2;
if(a[i][mid] < target) left = mid+1;
else if(a[i][mid] > target) right = mid-1;
else return true;
}
}
}
return false;
}
};
/*
* 方法二：缩小范围，从左下角元素（该行中最小数，该列中最大数）开始比较（利用L型有序）
* 过程：选取矩阵左下角元素，如果等于要查找的数，查找结束，
如果小于目标数，找下一列，j++
如果大于目标数，找上一行，i--
* 此方法效率较第一种高
O(r)或O(c)
*/
class Solution
{
public:
bool searchMatrix(vector<vector<int>>& a, int target)
{
if(a.empty() || a[0].empty()) return false; //若为空，则返回false
int row = a.size(); //行数
int col = a[0].size(); //列数
int i = row-1, j = 0;//从左下角元素开始比较
while(i >= 0 && j <= col-1)
{
if(a[i][j] < target) //如果小于，则找下一列
j++;
else if(a[i][j] > target) //如果大于，找上一行
i--;
else
return true;
}
return false;
}
};