Search a 2D Matrix (M)

題目

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 1:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 3
Output: true

Example 2:

Input:
matrix = [
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
target = 13
Output: false

題意

判斷在一個行元素遞增、列元素也遞增的矩陣中能否找到目標值，

思路

有序序列中找值那肯定是要用二分法，不過可以有兩種二分查找的方式：

1. 先一個二分找到目標值應該在的行，再一個二分在該行中查找目標值，
2. 由于矩陣的特殊性(行遞增，且下一行元素都比上一行元素大)，可以將矩陣直接當成長度為 m*n 的一維陣列處理，
3. 分治法，從左下角開始搜索，target更大則向右走，target更小則向上走(該題里實際就是暴力遍歷)，

代碼實作

Java

矩陣二分

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length;
        if (m == 0) return false;
        int n = matrix[0].length;
        if (n == 0) return false;

        if (target < matrix[0][0]) {
            return false;
        }

        // 先找行
        int left = 0, right = m - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (target < matrix[mid][0]) {
                right = mid - 1;
            } else if (target > matrix[mid][0]) {
                left = mid + 1;
            } else {
                return true;
            }
        }
        int r = left - 1;

        // 再找列
        left = 0;
        right = n - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (target < matrix[r][mid]) {
                right = mid - 1;
            } else if (target > matrix[r][mid]) {
                left = mid + 1;
            } else {
                return true;
            }
        }
        
        return false;
    }
}

一維二分

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        int m = matrix.length;
        if (m == 0) return false;
        int n = matrix[0].length;
        if (n == 0) return false;

        int left = 0, right = m * n - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            // 一維坐標轉換為二維坐標
            int x = mid / n;
            int y = mid % n;
            if (target < matrix[x][y]) {
                right = mid - 1;
            } else if (target > matrix[x][y]) {
                left = mid + 1;
            } else {
                return true;
            }
        }
        return false;
    }
}

JavaScript

矩陣二分

/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function (matrix, target) {
  if (matrix.length === 0 || matrix[0].length === 0) {
    return false
  }

  let m = matrix.length
  let n = matrix[0].length

  let left = 0
  let right = m - 1

  while (left <= right) {
    let mid = Math.trunc((right - left) / 2) + left
    if (matrix[mid][0] < target) {
      left = mid + 1
    } else if (matrix[mid][0] > target) {
      right = mid - 1
    } else {
      return true
    }
  }

  if (right >= 0) {
    let row = right
    left = 0
    right = n - 1

    while (left <= right) {
      let mid = Math.trunc((right - left) / 2) + left
      if (matrix[row][mid] < target) {
        left = mid + 1
      } else if (matrix[row][mid] > target) {
        right = mid - 1
      } else {
        return true
      }
    }
  }

  return false
}

分治法

/**
 * @param {number[][]} matrix
 * @param {number} target
 * @return {boolean}
 */
var searchMatrix = function (matrix, target) {
  if (matrix.length === 0 || matrix[0].length === 0) {
    return false
  }

  let m = matrix.length, n = matrix[0].length
  let i = m - 1, j = 0

  while (i >= 0 && j < n) {
    if (matrix[i][j] < target) {
      j++
    } else if (matrix[i][j] > target) {
      i--
    } else {
      return true
    }
  }

  return false
}

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