leetcode--Search a 2D Matrix

题目

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:1.Integers in each row are sorted from left to right.2.The first integer of each row is greater than the last integer of the previous row.

解法1：有序数组的折半查找
先找出target属于哪一行，然后在此行中进行查找

public boolean searchMatrix(int[][] matrix, int target) {        if(matrix.length == 0 || matrix[0].length == 0)            return false;        int start = 0, end = matrix.length-1;        if(target < matrix[start][0] || target > matrix[matrix.length-1][matrix[0].length-1])            return false;        if(target >= matrix[end][0])            start = end;        else{            while(end - start > 1){                int mid = (end + start)/2;                if(target >= matrix[mid][0] && target < matrix[end][0])                    start = mid;                else                    end = mid;            }        }        int row = start;        start = 0;        end = matrix[0].length-1;        while(start <= end){            int mid = (start+end)/2;            if(target < matrix[row][mid])                end = mid - 1;            else if(target > matrix[row][mid])                start = mid + 1;            else                return true;        }        return false;    }

解法2
将二维矩阵看作是一维数组，进行折半查找
对应关系：

m*n matrix[i][j] == array[i*n + j]
array[x] == matrix[x/n][x%n]
折半查找

public boolean searchMatrix(int[][] matrix, int target) {        if(matrix.length == 0 || matrix[0].length == 0)            return false;        int m = matrix.length,n = matrix[0].length;        int l = 0,r = m*n - 1;        while(l < r){            int x = (l + r)/2;            if(matrix[x/n][x%n] == target)                return true;            else if(matrix[x/n][x%n] <= target)                l = x + 1;            else                r = x - 1;        }        return matrix[l/n][l%n] == target;    }