LeetCode-74&240.Search a 2D Matrix

74 https://leetcode.com/problems/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.

For example,

Consider the following matrix:

[  [1,   3,  5,  7],  [10, 11, 16, 20],  [23, 30, 34, 50]]

Given target = 3, return true.

显然用二分查找

public bool SearchMatrix(int[,] matrix, int target)    {        int m = (int)matrix.GetLongLength(0);        int n = (int)matrix.GetLongLength(1);        if (matrix[0, 0] > target || matrix[m - 1, n - 1] < target)            return false;        int start = 0, end = m - 1, mid,row;        while (start<=end)        {            mid = (start + end) / 2;            if (matrix[mid, 0] == target)                return true;            if (matrix[mid, 0] < target)                start = mid + 1;            else                end = mid - 1;        }        row = end;        start = 0;        end = n - 1;        while (start<=end)        {            mid = (start + end) / 2;            if (matrix[row, mid] == target)                return true;            if (matrix[row, mid] < target)                start = mid + 1;            else                end = mid - 1;        }        return false;    }

240 https://leetcode.com/problems/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.

For example,

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]]

Given target = 5, return true.

Given target = 20, return false.

比较容易想到的办法及就是分别找行和列的max值

然后逐一判断

public static bool SearchMatrix(int[,] matrix, int target)        {            int m = (int)matrix.GetLongLength(0);            int n = (int)matrix.GetLongLength(1);            if (matrix[0, 0] > target || matrix[m - 1, n - 1] < target)                return false;            int l = 0, r = n - 1, mid=0,maxi,maxj;            while (l<=r)            {                mid = (l + r) / 2;                if (matrix[0, mid] == target)                    return true;                if (matrix[0, mid] > target)                    r = mid - 1;                else                    l = mid + 1;            }            maxj = r;            l = 0;            r = m - 1;            while (l <= r)            {                mid = (l + r) / 2;                if (matrix[mid,0] == target)                    return true;                if (matrix[mid,0] > target)                    r = mid - 1;                else                    l = mid + 1;            }            maxi = r;            for (int i = 0; i <= maxi; i++)            {                for (int j = 0; j <= maxj; j++)                {                    if (matrix[i, j] == target)                        return true;                }            }            return false;        }

第二种方法十分技巧，从左下角开始遍历, 如果大于当前就往右走, 小于就往上走。（也可以从右上角开始），参考https://segmentfault.com/a/1190000004288673

public bool SearchMatrix(int[,] matrix, int target)     {        int m = (int)matrix.GetLongLength(0);        int n = (int)matrix.GetLongLength(1);        if (matrix[0, 0] > target || matrix[m - 1, n - 1] < target)            return false;        int i = m - 1, j = 0;        while (i >= 0 && j < n)        {            if (matrix[i, j] == target)                return true;            if (matrix[i, j] > target)                i--;            else                j++;        }        return false;    }