[LeetCode]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.

思路：每行进行遍历，判断每行第一个是否小于目标，如果是则对该行进行二分查找，如果不是则说明这行以及下边所有行都不会存在目标，返回false

public class Solution {    public boolean searchMatrix(int[][] matrix, int target) {        if(matrix.length==0){            return false;        }        if(matrix[0].length==0){            return false;        }        for(int i=0;i<matrix.length;i++){            if(target<matrix[i][0]){                return false;            }            if(binarySearch(matrix[i],0,matrix[i].length-1,target)){                return true;            }        }        return false;    }        public boolean binarySearch(int[] nums,int start,int end,int target){        if(start>end){            return false;        }        int medium=(start+end)/2;        if(target==nums[start]||target==nums[end]||target==nums[medium]){            return true;        }else{            if(target>nums[medium]){                return binarySearch(nums,medium,end-1,target);            }else{                return binarySearch(nums,start+1,medium,target);            }                    }    }}