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]
]

Given target =

5

, return

true

.

Given target =

20

, return

false

.

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 row++;
}
return false;
}
}

binary search:

public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
int m = matrix.length;
if(m<1) return false;
int n = matrix[0].length;

return searchMatrix(matrix, new int[]{0,0}, new int[]{m-1, n-1}, target);
}

private boolean searchMatrix(int[][] matrix, int[] upperLeft, int[] lowerRight, int target) {
if(upperLeft[0]>lowerRight[0] || upperLeft[1]>lowerRight[1]
|| lowerRight[0]>=matrix.length || lowerRight[1]>=matrix[0].length)
return false;
if(lowerRight[0]-upperLeft[0]==0 && lowerRight[1]-upperLeft[1]==0)
return matrix[upperLeft[0]][upperLeft[1]] == target;
int rowMid = (upperLeft[0] + lowerRight[0]) >> 1;
int colMid = (upperLeft[1] + lowerRight[1]) >> 1;
int diff = matrix[rowMid][colMid] - target;
if(diff > 0) {
return searchMatrix(matrix, upperLeft, new int[]{rowMid, colMid}, target)
|| searchMatrix(matrix, new int[]{upperLeft[0],colMid+1}, new int[]{rowMid, lowerRight[1]}, target)
|| searchMatrix(matrix, new int[]{rowMid+1,upperLeft[1]}, new int[]{lowerRight[0], colMid}, target);
}
else if(diff < 0) {
return searchMatrix(matrix, new int[]{upperLeft[0], colMid+1}, new int[]{rowMid, lowerRight[1]}, target)
|| searchMatrix(matrix, new int[]{rowMid+1, upperLeft[1]}, new int[]{lowerRight[0], colMid}, target)
|| searchMatrix(matrix, new int[]{rowMid+1, colMid+1}, lowerRight, target);
}
else return true;
}
}

总结：思路很直接，利用binary search，可以理解为分治观点，但是时间复杂度比前一方法高。