leetcode 74 Search a 2D Matrix ----- java

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.

题目意思就是给定一个矩阵，按照从小到达排列，然后给定一个数字，判断数字是否在矩阵中。

就是用二分法的应用。两种方法，第一种稍微慢一些。

第一种是用了两次二分法，先找出行数、再在这一行上找具体是否存在。

第二种是直接应用二分法，效率更高。

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {


        int len1 = matrix.length;
        if( len1 == 0 )
            return false;
        int len2 = matrix[0].length;
        if( len2 == 0 || target < matrix[0][0])
            return false;
        
        int row_start = 0,row_end = len1-1;
        int col_start = 0,col_end = len2-1;
        int row = (row_end+row_start)/2 ,col;
        while( row_start <= row_end){
            row = (row_end+row_start)/2;
            if( target >= matrix[row][0] ){
                if( row == len1-1 || target < matrix[row+1][0])
                    break;
                else
                    row_start = row+1;
            }
            else{
                if( row == 0 || target >= matrix[row-1][0]){
                    row--;
                    break;
                }
                else
                    row_end = row-1;
            }
        }
        while( col_start <= col_end){
            col = (col_end+col_start)/2;

            if( target > matrix[row][col] ){
                if( col == col_end )
                    return false;
                else
                    col_start = col+1;
            }else if( target < matrix[row][col] ){
                if (col == col_start)
                    return false;
                else
                    col_end = col-1;
            }else
                return true;
        }
        return false;


    }
}

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {


        int len1 = matrix.length;
        if( len1 == 0 )
            return false;
        int len2 = matrix[0].length;
        if( len2 == 0 || target < matrix[0][0])
            return false;
        int start = 0,end = len1*len2-1,flag ;
        while( start <= end ){
            flag = (start+end)/2;
            int num = matrix[flag/len2][flag%len2];
            if( target > num)
                start = flag+1;
            else if ( target < num)
                end = flag-1;
            else
                return true;
        }
        return false;
        


    }
}