LeetCode 0074 - Search a 2D Matrix

Reck Zhang

发布于 2021-08-11 10:48:09

1590

发布于 2021-08-11 10:48:09

文章被收录于专栏：Reck Zhang

Search a 2D Matrix

Desicription

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.

Solution

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.empty() || matrix[0].empty())
            return 0;
        vector<int> col(matrix.size());
        for(int i = 0; i < matrix.size(); ++i)
            col[i] = matrix[i][0];
        int index = lower_bound(col.begin(), col.end(), target) - col.begin();
        if(!index && col[index] > target)
            return 0;
        if(index == matrix.size() ||(index && matrix[index][0] > target))
            --index;
        return *lower_bound(matrix[index].begin(), matrix[index].end(), target) == target;
    }
};

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原始发表：2017-11-19，如有侵权请联系 cloudcommunity@tencent.com 删除

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