Search a 2D Matrix 题解
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题目来源:https://leetcode.com/problems/search-a-2d-matrix/description/
Description
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.
Example
For example,
Consider the following matrix:
[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Given target = 3
, return true
.
Solution
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if (matrix.empty() || matrix[0].empty()) {
return false;
}
int size = matrix.size();
int low = 0, high = size - 1, mid;
while (low < high) {
mid = (high + low) / 2;
if (target == matrix[mid].back())
return true;
else if (target < matrix[mid].back())
high = mid;
else
low = mid + 1;
}
size = matrix[low].size();
vector<int>& arr = matrix[low];
low = 0;
high = size - 1;
while (low < high) {
mid = (high + low) / 2;
if (target == arr[mid])
return true;
else if (target < arr[mid])
high = mid;
else
low = mid + 1;
}
return arr[low] == target;
}
};
解题描述
这道题考察的是二分查找。我选择的算法是先用二分查找确定target在矩阵的哪一行,再在这一行中进行二分查找找出target。不过AC之后想想,其实可以把矩阵直接看成一维数组进行二分查找,只需要多做一步下标转换即可。