LeetCode 240. Search a 2D Matrix II

最新推荐文章于 2025-01-17 23:23:30 发布

原创最新推荐文章于 2025-01-17 23:23:30 发布 · 164 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 题解专栏收录该内容

288 篇文章

订阅专栏

本文介绍了一种高效的矩阵搜索算法，该算法能在排序矩阵中快速查找目标值。矩阵的每一行从左到右递增排序，每一列从上到下递增排序。通过观察，可以发现从矩阵的最后一行第一列开始搜索，若当前值小于目标值则向右移动，大于目标值则向上移动，直至找到目标值或超出边界。

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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

Observe carefully... 0 colomn and m-1 row is a increasing sequence.... same with 1 column and m-2 row....etc al.

#include <vector>
#include <iostream>
using namespace std;

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

int main(void) {
    vector< vector<int> > matrix{
        {1, 4, 7, 11, 15},
        {2, 5, 8, 12, 19},
        {3, 6, 9, 16, 22},
        {10, 13, 14, 17, 24},
        {18, 21, 23, 26, 30}};
    bool found = searchMatrix(matrix, 5);
    cout << found << endl;
}