[LeetCode92]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.

Analysis:

According to the rule of the matrix, 

• 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

The minimum number is on the left top, the maximum number is on the right bottom.

so we begin search on the right top, if target>matrix[i][j] move down, else move left.

This algorithm taks O(m+n)

c++

bool searchMatrix(vector<vector<int> > &matrix, int target) {    int row = matrix.size();    if(row<=0) return false;    int col = matrix[0].size();    if(col<=0) return false;    if(target<matrix[0][0] || target> matrix[row-1][col-1]) return false;    int x1 = 0;    int y1 = col-1;    while(x1<=row-1 && y1>=0){        if(target==matrix[x1][y1])            return true;        else if(target>matrix[x1][y1]){            x1++;        }else{            y1--;        }    }    return false;}

java

public boolean searchMatrix(int[][] matrix, int target) {int row = matrix.length;    if(row<=0) return false;    int col = matrix[0].length;    if(col<=0) return false;    if(target<matrix[0][0] || target> matrix[row-1][col-1]) return false;    int x1 = 0;    int y1 = col-1;    while(x1<=row-1 && y1>=0){        if(target==matrix[x1][y1])            return true;        else if(target>matrix[x1][y1]){            x1++;        }else{            y1--;        }    }    return false;}