Search a 2D Matrix(medium)

【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.

【题意】

       在一个二维矩阵中找到给定的值。矩阵从上到下从左到右有序

【分析】

    与剑指offer里面的一个题类似，思路是从右上角元素开始遍历，每次遍历中若与target相等则返回true；若小于则行向下移动；若大于则列向左移动。时间复杂度m+n

【实现】

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;    }};

参考别人：

     第二种方法：二分法确定target可能在第几行出现。再用二分法在该行确定target可能出现的位置。时间复杂度O(logn+logm)

class Solution {public:    bool searchMatrix(vector<vector<int> > &matrix, int target) {        int left = 0;        int right = matrix.size()-1;        if(left != right)        {            while(left<=right)            {                int middle = left + (right-left)/2;                if(matrix[middle][0] < target)                {                    left = middle+1;                }                else if(matrix[middle][0] > target)                {                    right = middle-1;                }                else                {                    return true;                }            }        }        if(right == -1)        {            return false;        }        else        {            int row = right;            int left = 0;            int right = matrix[row].size()-1;            while(left<=right)            {                int middle = left + (right-left)/2;                if(matrix[row][middle] < target)                {                    left = middle+1;                }                else if(matrix[row][middle] > target)                {                    right = middle-1;                }                else                {                    return true;                }            }            return false;        }            }};