240. Search a 2D Matrix II

题目：

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]
]

Given target = 5, return true.

Given target = 20, return false.

链接：https://leetcode.com/problems/search-a-2d-matrix-ii/#/description

4/27/2017

53%, 15ms

从右上和左下两个方向进行，缩小范围到left/right重合或者交叉，或者top/bottom重合交叉。

注意第25-27行没有的话对结果也没有影响，但是缺少的话最后一个else其实是包含了很多非法情况的。

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        
        int right = matrix[0].length - 1, left = 0, top = 0, bottom = matrix.length - 1;
        while (left < right && top < bottom) {
            if (matrix[top][right] < target) {
                top++;
            } else if (matrix[top][right] > target) {
                right--;
            } else {
                return true;
            }
            if (matrix[bottom][left] < target) {
                left++;
            } else if (matrix[bottom][left] > target) {
                bottom--;
            } else {
                return true;
            }
        }
        
        if (right < left || bottom < top) {
            return false;
        }
        if (right == left) {
            while (top + 1 < bottom) {
                int mid = top + (bottom - top) / 2;
                if (matrix[mid][right] == target) {
                    return true;
                } else if (matrix[mid][right] > target) {
                    bottom = mid;
                } else {
                    top = mid;
                }
            }
            if (matrix[top][right] == target || matrix[bottom][right] == target) {
                return true;
            } else {
                return false;
            }
        } else {
            while (left + 1 < right) {
                int mid = left + (right - left) / 2;
                if (matrix[top][mid] == target) {
                    return true;
                } else if (matrix[top][mid] > target) {
                    right = mid;
                } else {
                    left = mid;
                }
            }
            if (matrix[top][right] == target || matrix[top][left] == target) {
                return true;
            } else {
                return false;
            }            
        }
    }
}

别人的算法：

只需要从右上开始，比target大就是right--，比target小就top++，直到找到target或者超出边界

https://discuss.leetcode.com/topic/20064/my-concise-o-m-n-java-solution

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

很有意思的divide-conquer方法，虽然效率并不高。时间复杂度并不是评论中所说的那么少。

https://discuss.leetcode.com/topic/33240/java-an-easy-to-understand-divide-and-conquer-method

下面是关于这个解法的时间复杂度的讨论，可以看看最后2个链接

https://discuss.leetcode.com/topic/33240/java-an-easy-to-understand-divide-and-conquer-method/15

https://discuss.leetcode.com/topic/21030/is-there-s-a-o-log-m-log-n-solution-i-know-o-n-m-and-o-m-log-n

http://stackoverflow.com/questions/2457792/given-a-2d-array-sorted-in-increasing-order-from-left-to-right-and-top-to-bottom/2458113#2458113

更多讨论：

https://discuss.leetcode.com/category/301/search-a-2d-matrix-ii