leetcode解题方案--074--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:

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.

分析

搜索一个矩阵中是否存在某个数

首先搜索所在行 首位指针二分法

然后搜索该行中有没有这个数

还有一种简单的方法，是依次遍历。先按照行查找，如果大就减一行，如果小久加j

public boolean searchMatrix(int[][] matrix, int target) {
if(matrix == null || matrix.length ==0 || matrix[0].length == 0) {
return false;
}

int i = matrix.length-1;
int j =0;
while(i >= 0 && j <= matrix[0].length -1) {
if(matrix[i][j] == target) {
return true;
} else if(matrix[i][j] > target) {
i --;
} else {
j++;
}
}
return false;
}

class Solution {
public static boolean searchMatrix(int[][] matrix, int target) {
if (matrix.length == 0 || matrix[0].length == 0) {
return false;
}

int low = 0;
int high = matrix.length-1;
int targetLine = 0;
if (target < matrix[low][0]) {
return false;
}
if (target >= matrix[high][0]) {
targetLine = high;
} else {
while (high-low>1) {
int mid = low+(high-low)/2;
if (matrix[mid][0]>target) {
high = mid;
} else {
low = mid;
}
}
targetLine = low;
}
low = 0;
high = matrix[targetLine].length-1;
if (matrix[targetLine][low] == target || target==matrix[targetLine][high]) {
return true;
}

while (high-low>1) {
int mid = low+(high-low)/2;
if (matrix[targetLine][mid]>target) {
high = mid;
} else {
low = mid;
}
if (matrix[targetLine][low] == target || target==matrix[targetLine][high]) {
return true;
}

}
return false;
}
}