LeetCode解题报告（300)-- 240. Search a 2D Matrix II

Problem

Write an efficient algorithm that searches for a target value in an m x n integer matrix. The 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.

Example 1:

Input: 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]], target = 5
Output: true

Example 2:

Input: 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]], target = 20
Output: false

Constraints:

• m == matrix.length
• n == matrix[i].length
• 1 <= n, m <= 300
• -109 <= matix[i][j] <= 109
• All the integers in each row are sorted in ascending order.
• All the integers in each column are sorted in ascending order.
• -109 <= target <= 109

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Analysis

  这是一道系列题，上一道题目的矩阵是按行递增，但这里有不一样，它是向右、下两个方向递增的，但是这两个方向并没有说谁大谁小，所以搜索起来会比较麻烦。这道题目巧妙的地方在于遍历的起点并不是左上角，而是左下角（或右上角）。我们观察Example 1中的左下角元素18，比它大的元素都在它的右侧，比它小的元素都在它的上方。这里就是我们要找的规律，对于矩阵中的元素来说，它的上方一定比它小，而它的右方一定比它大，所以我们就借助这个特点，从左下角开始遍历。如果越界说明找不到元素了，直接return false即可。z

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Solution

  无

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Code

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

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Summary

  这道题目巧妙的地方在于它遍历的起点是左下角，这与常规从左上角遍历是很不一样的，所以当题目给出的矩阵有一定的特点之后，我不妨从四个顶点都试一下，看看有没有一些可以利用的性质。这道题目的分享到这里，谢谢您的支持！