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。