search in a 2D array.png
===================== 解題思路 =====================
同樣以 BS 的做法來找 搜索範(fàn)圍是 0 ~ mn -1 (總共就 mn 個(gè)數(shù)值), 設(shè)定 left right 跟 mid 每次減半搜索範(fàn)圍, 但題目給的是一個(gè) 2D 陣列, 所以要將行與列來做一個(gè)轉(zhuǎn)換拿到真正在 2D 陣列裡的位置 同樣這類型題目都採(cǎi)用 while(left + 1 < right) 的做法避免死循環(huán), 所以在結(jié)束 loop 以後要對(duì) left 跟 right 各自檢查一次是否符合題目需要
===================== C++ code ====================
<pre><code>
class Solution {
public:
/**
* @param matrix, a list of lists of integers
* @param target, an integer
* @return a boolean, indicate whether matrix contains target
*/
bool searchMatrix(vector<vector<int> > &matrix, int target) {
// write your code here
if(matrix.size() == 0) return false;
int m = matrix.size(), n = matrix[0].size();
int left = 0, right = m * n - 1;
while(left + 1 < right)
{
int mid = left + (right - left) / 2;
if(matrix[mid / n][mid % n] == target) return true;
else if(matrix[mid / n][mid % n] > target) right = mid;
else left = mid;
}
if(matrix[left / n][left % n] == target) return true;
if(matrix[right / n][right % n] == target) return true;
return false;
}
};
<code><pre>
