[LeetCode]Search a 2D Matrix

题目

Number: 74
Difficulty: Medium
Tags: Array, Binary Search

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.

题解

在一个排序矩阵中寻找给定值。

代码

用两次二分查找，在第一列中找到给定值的下届，再在这一行中查找给定值：16ms

bool searchMatrix(vector<vector<int>>& matrix, int target) {
    int m = matrix.size(), n = matrix[0].size();
    int l = 0, h = m - 1;
    if(matrix[0][0] > target || matrix[m - 1][n - 1] < target)
        return false;
    while(l < h){
        int mid = l + (h - l + 1) / 2;
        if(matrix[mid][0] > target)
            h = mid - 1;
        else
            l = mid;
    }
    int lrow = 0, hrow = n - 1;
    while(lrow <= hrow){
        int mid = lrow + (hrow - lrow) / 2;
        if(matrix[l][mid] > target)
            hrow = mid - 1;
        else if(matrix[l][mid] < target)
            lrow = mid + 1;
        else
            return true;
    }
    return false;
}

可以把矩阵当做数组直接用一次二分查找：16ms

bool searchMatrix(vector<vector<int>>& matrix, int target) {
    int m = matrix.size(), n = matrix[0].size();
    int l = 0, h = m * n - 1;
    if(matrix[0][0] > target || matrix[m - 1][n - 1] < target)
        return false;
    while(l <= h){
        int mid = l + (h - l) / 2;
        int res = matrix[mid / n][mid % n];
        if(res > target)
            h = mid - 1;
        else if(res < target)
            l = mid + 1;
        else
            return true;
    }
    return false;
}

参考《剑指Offer》面试题3，每次循环排除一列或者一行：12ms

bool searchMatrix(vector<vector<int>>& matrix, int target) {
    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)
            j--;
        else
            i++;
    }
    return false;
}