Search a 2D Matrix II

ID: 38; medium; 搜索二维矩阵（二）

LintCode 炼码

Solution 1 (Java)

public class Solution {
    /**
     * @param matrix: A list of lists of integers
     * @param target: An integer you want to search in matrix
     * @return: An integer indicate the total occurrence of target in the given matrix
     */
    public int searchMatrix(int[][] matrix, int target) {
        int row = 0, col = matrix[0].length - 1;
        int targetCount = 0;
        while (row < matrix.length && col >= 0) {
            if (matrix[row][col] == target) {
                targetCount++;
                row++;
                col--;
            } else if (matrix[row][col] < target) {
                row++;
            } else {
                col--;
            }
        }
        return targetCount;
    }
}

Notes

• We pick a special number in the 2D matrix here. Either the top right element or the bottom left element will work. The reason is that we want to reduce the range in which we research for the target. In the code above, we chose the top right element and it is the largest in its row but the smallest in its column. Thus, if the target is larger than it, we proceed downward to find it; if the target is smaller than it, we proceed to the left. And if the target is equal to it, we increment the targetCounter and move diagonally since each number is unique in its row and column.