Matrix

1. Search a 2D Matrix

2. Search a 2D Matrix II

1. Search a 2D Matrix

1. O(log(row) + log (col)): 2 x binary search
2. O(log (row * col)): flatten 2D matrix to 1D array

• 2 x binary search

需要注意的是在进入row search的时候的判断条件：matrix[end][0] <= target

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;

        int start = 0, end = matrix.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (matrix[mid][0] == target) {
                return true;
            } else if (matrix[mid][0] < target) {
                start = mid;
            } else {
                end = mid;
            }
        }
        if (matrix[end][0] <= target) {
            return searchRow(matrix, end, target);
        } else {
            return searchRow(matrix, start, target);
        }
    }

    private boolean searchRow(int[][] matrix, int row, int target) {
        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;
            } else if (matrix[row][mid] < target) {
                left = mid;
            } else {
                right = mid;
            }
        }
        if (matrix[row][left] == target || matrix[row][right] == target) return true;
        return false;
    }
}

• flatten 2D matrix to 1D array

题目中给出的两个条件可以让整个matrix变成一个sorted array。

注意：如何转换成row, col

col = mid % matrix[0].length;
row = mid / matrix[0].length;

public boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
    int start = 0, end = matrix.length * matrix[0].length - 1;
    int col = 0, row = 0;
    while (start + 1 < end) {
        int mid = start + (end - start) / 2;
        col = mid % matrix[0].length;
        row = mid / matrix[0].length;

        if (matrix[row][col] == target) {
            return true;
        } else if (matrix[row][col] < target) {
            start = mid;
        } else {
            end = mid;
        }
    }
    int num1 = matrix[start / matrix[0].length][start % matrix[0].length];
    int num2 = matrix[end / matrix[0].length][end % matrix[0].length];
    if (num1 == target || num2 == target) return true;
    return false;
}

2. Search a 2D Matrix II

从左下角开始走

O(row + col)，最多是走到了对角线的最后一个点。每次row和col都是走一格子，最多是row + col。