74. Search a 2D Matrix

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 from left to right.
• The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[  [1,   3,  5,  7],  [10, 11, 16, 20],  [23, 30, 34, 50]]

Given target = 3, return true.

这道题目主要涉及二分查找和数组的知识。关于二分查找，之前有过类似的题目，这个算法很经典，主要要注意left和right的更新（需要和判断条件对应）：有以下几种情况等价：

1. if(mid<tar) right=mid-1;   else left=mid;

2.if(mid=tar) return mid; else if(mid<tar) right=mid-1;  else right=mid+1;

3.if(mid>tar) left=mid+1; else right=mid;

这道题因为已经对数组排好序了，所以直接先对行搜索，如果找到所在的行就在这一行二分搜索。

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