Search a 2D Matrix II.java

M
1528169466
tags: Binary Search, Divide and Conquer

给matrix, 每一行sorted, 每一列从上往下sorted, 找target是否存在

#### Binary Search
- 根据给定的性质, 其实点选的极端一点: x = 最下面的row, y = 当下一行里面最小的left position. 
- (x,y)在左下角
- 在此情况下, 只能往一个方向运行: 如果小于target, y++; 如果大于target, 那么只能x--
- 每次操作, 都是删掉一行, 或者一列, 再也不需要回头看
- `while (x >= 0 && y < col) {}` 确保不会跑脱
- 同样的方式: 可以从右上角(0, col - 1) 开始, 代码稍微改一改

#### Divide and Conquer?
- TODO

```
/**
LeetCode: check existance
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.
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]
]
Example 1:

Input: matrix, target = 5
Output: true
Example 2:

Input: matrix, target = 20
Output: false
*/
// from bottom-left, (x,y) = (row -1, 0)
public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        int row = matrix.length;
        int col = matrix[0].length;
        int x = row - 1;
        int y = 0;
        
        while (x >= 0 && y < col) {
            if (matrix[x][y] < target) {
                y++;
            } else if (matrix[x][y] > target) {
                x--;
            } else {//matrix[x][y] == target
                return true;
            }
        }
        return false;
    }
}

// from top-right, (x,y) = (0, col - 1)
public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        int row = matrix.length;
        int col = matrix[0].length;
        int x = 0;
        int y = col - 1;
        
        while (x < row && y >= 0) {
            if (matrix[x][y] < target) {
                x++;
            } else if (matrix[x][y] > target) {
                y--;
            } else {//matrix[x][y] == target
                return true;
            }
        }
        return false;
    }
}

/*
LintCode: find # of occurances
Write an efficient algorithm that searches for a value in an m x n matrix, return the occurrence of it.

This matrix has the following properties:

    * Integers in each row are sorted from left to right.

    * Integers in each column are sorted from up to bottom.

    * No duplicate integers in each row or column.

Example
Consider the following matrix:

[

    [1, 3, 5, 7],

    [2, 4, 7, 8],

    [3, 5, 9, 10]

]

Given target = 3, return 2.

Challenge
O(m+n) time and O(1) extra space

*/

/*
    Thoughts: 11.29.2015. LintCode updated. Practice again.
    Understand that:
    1. Each row has exactly 1 instance of that integer, no duplicates.
    2. Each row begins with smallest number. So, if matrix[x][y] < target, first thing to do is x++.
    3. Each col ends with largest number. So if matrix[x][y] > target, 
        (no need to care x++ since it's alreay too large for this row), then simply just y--.
    4. If match, next step will be x--,y++. 
        x-- because it has to change a row; 
        y++ because [x-1, y] can't be the target since no duplicate in column
    Beneftis of going from bottown-left: No matter which condition, always have 1 possible way to move.
*/

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



```