Search a 2D Matrix

标签：leetcode

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.

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length;
        if(m==0)
            return false;
        int n=matrix[0].length;
        if(matrix[m-1][n-1]<target||matrix[0][0]>target)
            return false;
            
        int first=0;
        int last=m-1;
        //二分法,判断target所在的行
        //判断条件不能使<=,防止m==1时,while是死循环
        while(first<last){
            int center=(first+last)/2;
            if(matrix[center][n-1]<target){
                first=center+1;
            }else if(matrix[center][n-1]==target){
                return true;
            }
            else{
                last=center;
            }
        }
        
        int left=0;
        int right=n-1;
        //二分法,判断target所属的列
        while(left<=right){
            int center=(left+right)/2;
            if(matrix[first][center]==target)
                return true;
            else if(matrix[first][center]<target){
                left=center+1;
                
            }else{
                right=center-1;
            }
        }
        return false;
    }
}

Search a 2D Matrix

标签：leetcode

原文地址：http://blog.csdn.net/u010786672/article/details/41985241