[LintCode] 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.

4/23/2017

算法班

corner case想的不够全面，第23-28行是后来加上去的，原因是如果target在最后一行或者比任何一个值都大，我们是需要在end那一行找答案的。其他所有情况都是在start那一行找。

public class Solution {
    /**
     * @param matrix, a list of lists of integers
     * @param target, an integer
     * @return a boolean, indicate whether matrix contains target
     */
    public boolean searchMatrix(int[][] matrix, int target) {
        // write your code here
        
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
        boolean ret;
        int start = 0, end = matrix.length - 1;
        
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (matrix[mid][0] == target) return true;
            if (matrix[mid][0] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        int row;
        if (matrix[end][0] < target) {
            row = end;
        } else {
            row = start;
        }
        int left = 0, right = matrix[row].length - 1;
        while (left + 1 < right) {
            int mid = left + (right - left) / 2;
            if (matrix[row][mid] == target) return true;
            if (matrix[row][mid] < target) {
                left = mid;
            } else {
                right = mid;
            }
        }
        if (matrix[row][right] == target || matrix[row][left] == target) return true;
        return false;
    }
}