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leetcode 074 —— Search a 2D Matrix

2015-07-31 15:37 309 查看

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 =

, return

true

.

思路：从左上角或者右下角的顶点开始搜索

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int m = matrix.size();
int n = matrix[0].size();
int i = 0, j = n - 1;
while (i<m&&j >= 0){
if (matrix[i][j] == target)
return true;
else if (matrix[i][j]>target)
j--;
else
i++;
}
return false;

}
};

第二种方法：最优路径搜索，也能通过，16ms，但是程序有待优化

class Solution {
public:
 bool searchMatrix(vector<vector<int>>& matrix, int target) {
 vector<vector<int>> &a = matrix;
 int m = a.size();
 int n = a[0].size();
 int i = 0, j = 0;
 while (i<m&&j<n){
 if (a[i][j]>target) return false;
 if (a[i][j] == target) return true;
 if (i == m - 1){
 while (j < n){
 if (a[i][j] == target) return true;
 else if (a[i][j] < target) j++;
 else
 return false;
 }
 return false;
 }
 if (j == n - 1){
 while (i < m){
 if (a[i][j] == target) return true;
 else if (a[i][j] < target) i++;
 else
 return false;
 }
 return false;
 }
 if (a[i + 1][j] < target&&a[i][j + 1] < target){
 if (a[i + 1][j]>a[i][j + 1]) i++;
 else j++;
 continue;
 }
 if (a[i + 1][j]>target&&a[i][j + 1]>target) return false;
 if (a[i + 1][j] == target || a[i][j + 1] == target) return true;
 if (a[i + 1][j]<target&&a[i][j + 1]>target){
 i++;
 continue;
 }
 if (a[i + 1][j] > target&&a[i][j + 1] < target){
 j++;
 continue;
 }
 }
 }
}a;

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