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面试题 04. 二维数组中的查找

题目描述

在一个 n * m 的二维数组中,每一行都按照从左到右递增的顺序排序,每一列都按照从上到下递增的顺序排序。请完成一个函数,输入这样的一个二维数组和一个整数,判断数组中是否含有该整数。

示例:

现有矩阵 matrix 如下:

[
  [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]
]

给定 target = 5,返回  true

给定  target = 20,返回  false

限制:

  • 0 <= n <= 1000

  • 0 <= m <= 1000

解法

从左下角(或右上角)开始查找即可。

Python3

class Solution:
    def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
        if not matrix or not matrix[0]:
            return False
        rows, cols = len(matrix), len(matrix[0])
        i, j = rows - 1, 0
        while i >= 0 and j < cols:
            if matrix[i][j] == target:
                return True
            if matrix[i][j] > target:
                i -= 1
            else:
                j += 1
        return False

Java

class Solution {
    public boolean findNumberIn2DArray(int[][] matrix, int target) {
        int m, n;
        if (matrix == null || (m = matrix.length) == 0 || matrix[0] == null || (n = matrix[0].length) == 0) return false;
        int i = 0, j = n - 1;
        while (i < m && j >= 0) {
            if (matrix[i][j] == target) return true;
            if (matrix[i][j] > target) --j;
            else ++i;
        }
        return false;
    }
}

JavaScript

/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var findNumberIn2DArray = function (matrix, target) {
  let m, n;
  if (
    matrix == null ||
    (m = matrix.length) == 0 ||
    matrix[0] == null ||
    (n = matrix[0].length) == 0
  )
    return false;
  let i = 0,
    j = n - 1;
  while (i < m && j >= 0) {
    if (matrix[i][j] == target) return true;
    if (matrix[i][j] > target) --j;
    else ++i;
  }
  return false;
};

Go

func findNumberIn2DArray(matrix [][]int, target int) bool {
    if len(matrix) == 0 {
        return false
    }
    rows, cols := len(matrix), len(matrix[0])
    i, j := rows - 1, 0
    for i >= 0 && j < cols {
        if matrix[i][j] == target {
            return true
        }
        if matrix[i][j] > target {
            i--
        } else {
            j++
        }
    }
    return false
}

C++

class Solution {
public:
    bool findNumberIn2DArray(vector<vector<int>>& matrix, int target) {
        if (matrix.empty()) {
            return false;
        }
        int m = matrix.size(), 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) {
                ++i;
            } else {
                --j;
            }
        }
        return false;
    }
};

TypeScript

function findNumberIn2DArray(matrix: number[][], target: number): boolean {
    let m: number = matrix.length, n: number;
    if (!matrix || !m || !matrix[0] || !(n = matrix[0].length)) return false;
    let i: number = 0, j: number = n - 1;
    while (i < m && j >= 0) {
        let cur: number = matrix[i][j];
        if (cur == target) return true;
        if (cur > target) {
            j--;
        } else {
            i++;
        }
    }
    return false;
};

...

Java
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