[Leetcode]Search a 2D matrix

[timu]

Search a 2D Matrix

Total Accepted: 52191 Total Submissions: 164580

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.

[silu]

in fact, I think it is a 2d binary search, first binary search line and locate the line.

second, binary search the line and find the target.

[daima]

my accepted solution:

    public boolean searchMatrix(int[][] matrix, int target) {        if(matrix == null || matrix.length == 0) return false;        int left = 0;        int right = matrix.length - 1;        while(left <= right){            int mid = left+ (right - left)/2;            int begin = matrix[mid][0];            int end = matrix[mid][matrix[0].length - 1];            if(target < begin){                right = mid - 1;            }else if(target > end){                left = mid + 1;            }else{                return binarySearch(target, matrix[mid]);            }        }        return false;    }    private boolean binarySearch(int target, int[] nums){        if(nums==null|| nums.length == 0) return false;        int start = 0;        int end = nums.length - 1;        while(start<=end){            int mid = start + (end - start)/2;            if(target == nums[mid]) return true;            else if(target > nums[mid]){                start = mid + 1;            }else{                end = mid - 1;            }        }        return false;    }    

or zhijie fa:

public boolean searchMatrix(int[][] matrix, int target) {    int row_num = matrix.length;    int col_num = matrix[0].length;    int begin = 0, end = row_num * col_num - 1;    while(begin <= end){        int mid = (begin + end) / 2;        int mid_value = matrix[mid/col_num][mid%col_num];        if( mid_value == target){            return true;        }else if(mid_value < target){            //Should move a bit further, otherwise dead loop.            begin = mid+1;        }else{            end = mid-1;        }    }    return false;}