[leetcode 240/74] Search a 2D Matrix II -----在左右有序，上下有序中查找数据

[leetcode 240]题目：

Quesiton:

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.

[leetcode 74]题目：

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.

分析：

从右上角第一个数据比较开始即可，如果这个数据比右上角数据大，则向下走，如果比右上角数据小则向左走即可，相等就返回。

代码如下：

<span style="font-size:14px;">class Solution {public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {        if(matrix.size() == 0 || matrix[0].size() == 0 || target < matrix[0][0])            return false;        int left = 0;        int bottom = matrix.size();        int row = 0;        int col = matrix[0].size()-1;        while(row < bottom && col >= left){            if(target == matrix[row][col])                return true;            else                if(target < matrix[row][col]){                    --col;                }                else{                    ++row;                }        }        return false;    }};</span>