[LeetCode] Search a 2D Matrix

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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.

解答：

class Solution {public:    bool searchMatrix(vector<vector<int> > &matrix, int target) {        if(matrix.size() == 0 || matrix[0].size() == 0) {            return false;        }                int m = matrix[0].size();        int n = matrix.size();        if(target < matrix[0][0] || target > matrix[n-1][m-1]) {            return false;        }        int head, cur_1, cur_2, end;        //第一轮二分查找        head = 0;        end = n-1;        while(head <= end) {            cur_1 = (head+end)/2;            if(matrix[cur_1][0] == target) {                return true;            }            else if(matrix[cur_1][0] > target) {                end = end - 1;            }            else if(matrix[cur_1][0] < target) {                head = head + 1;            }        }        if(target < matrix[cur_1][0]) {            cur_1--;  //由此得到的cur，即为第二轮二分查找的那一行        }        //第二轮二分查找        head = 0;        end = m -1;        while(head <= end) {            cur_2 = (head+end)/2;            if(matrix[cur_1][cur_2] == target) {                return true;            }            else if(matrix[cur_1][cur_2] > target) {                end = end - 1;            }            else if(matrix[cur_1][cur_2] < target) {                head = head + 1;            }        }        return false;    }};

思路：

我的做法是两次二分查找，先确定元素在哪一行，然后在这一行再进行二分查找。这种算法的效率不低，但是编码稍微有些困难。网上有一个经典的算法，虽然效率没有我这种算法高，但是简单易懂，便于编写：

class Solution { public:     bool searchMatrix(vector<vector<int> > &matrix, int target) {         // Start typing your C/C++ solution below         // DO NOT write int main() function         int i = 0, j = matrix[0].size() - 1;                  while (i < matrix.size() && j >= 0)         {             if (target == matrix[i][j])                 return true;             else if (target < matrix[i][j])                 j--;             else                 i++;         }                  return false;     } };

还有一种做法是全局二分查找，即把二维数组当成一个一维数组，但是那种方法对下标的操作很复杂，不推荐用该做法解题。

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