240. Search a 2D Matrix II

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 in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Giventarget=5, returntrue.

Giventarget=20, returnfalse.

解题方法

T(n) = 3T(n/4) + O(1)

1. Recursive : Divide and conquer 透過劃分範圍
2. While loop: 移動pointer

//Divide and Conquer
    public boolean searchMatrix(int[][] matrix, int target){
        if(matrix == null || matrix.length == 0){
            return false;
        }
        if(matrix[0] == null || matrix[0].length == 0){
            return false;
        }
        int row = matrix.length;
        int col = matrix[0].length; 
        return searchMatrix(matrix, target, 0, 0, row - 1, col -1);
    }


    private boolean searchMatrix(int[][] matrix, int target, int startX, int startY, int endX, int endY){
        //ending condition
        if(endX < startX || endY < startY){
            return false;
        }

        int middleX = startX + (endX - startX) / 2;
        int middleY = startY + (endY -startY) / 2;
        if(matrix[middleX][middleY] == target)
        {
            return true;
        }else if(matrix[middleX][middleY] < target){
            return searchMatrix(matrix, target, startX, middleY + 1, endX, endY) || 
                searchMatrix(matrix, target, middleX + 1, startY, endX, middleY);
        }else{
            return searchMatrix(matrix, target, startX, startY, endX, middleY - 1) ||
                searchMatrix(matrix, target, startX, middleY, middleX - 1, endY);
        }
    }

//Moving pointer
    /*public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0){
            return false;
        }
        if(matrix[0] == null || matrix[0].length == 0){
            return false;
        }

        int row = matrix.length;
        int col = matrix[0].length;
        int x = 0;                        //**********右上角 或 左上角都可
        int y = col - 1;
        while( x <= row && y >= 0){
            if(matrix[x][y] == target){
                return true;
            }else if(matrix[x][y] < target){ //往大的移動
                x++;
            }else{
                col--;
            }
        }

        return matrix[x][y] == target;
    }*/