74. 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.

本题最笨的方法就是遍历所有的元素，代码如下：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix==null||matrix.length==0||matrix[0].length==0) return false;
        int row = matrix.length;
        int col = matrix[0].length;
        for(int i=0;i<row;i++){
            if(i<row-1&&target>=matrix[i+1][0]) continue;
            for(int j=0;j<col;j++){
                if(target>matrix[i][j]) continue;
                else if(target==matrix[i][j]) return true;
                else return false;
            }
        }
        return false;
    }
}

看了答案才知道，可以用binary search，代码如下：

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix==null||matrix.length==0) return false;
        int left = 0;
        int m = matrix.length;
        int n = matrix[0].length;
        int right = m*n-1;
        while(left<=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-1;
            else left = mid+1;
        }
        return false;
    }
}