LeetCode（74） Search a 2D Matrix

题目

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.
For example,

Consider the following matrix:

[
[1, 3, 5, 7],
[10, 11, 16, 20],
[23, 30, 34, 50]
]
Given target = 3, return true.

分析

二分搜索算法在二维矩阵中的应用；

两次二分搜索：

第一次，二分搜索第一列（借助另一个数组）找出小于等于target的下标pos

第二次，二分搜索pos行，搜索target

AC代码

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if (matrix.empty())
            return false;

        int m = matrix.size();

        //将矩阵首列，压入临时vector，寻找，目标元素所在的行
        vector<int> v;
        for (int i = 0; i < m; i++)
            v.push_back(matrix[i][0]);

        //寻找 <=target 最近的元素下标
        int pos = binarySearchPos(v, 0, m - 1, target);

        //不存在任何一行中
        if (pos == -1)
            return false;
        else if(matrix[pos][0] == target)
            return true;
        else
            return binarySearch(matrix[pos], 0, matrix[pos].size() - 1, target);

    }

    int binarySearchPos(vector<int> &nums, int lhs, int rhs, int &target)
    {
        if (nums.empty())
            return -1;

        while (lhs <= rhs)
        {
            int mid = (lhs + rhs) / 2;

            if (nums[mid] == target)
                return mid;

            else if (nums[mid] < target)
            {
                lhs = mid + 1;
            }
            else{
                rhs = mid - 1;
            }//else
        }//while

        if (lhs < rhs && nums[lhs] < target)
        {
            return lhs;
        }
        else{
            return lhs - 1;
        }
    }

    bool binarySearch(vector<int> &nums, int lhs, int rhs, int &target)
    {
        if (nums.empty())
            return false;

        while (lhs <= rhs)
        {
            int mid = (lhs + rhs) / 2;

            if (nums[mid] == target)
                return true;

            else if (nums[mid] < target)
            {
                lhs = mid + 1;
            }
            else{
                rhs = mid - 1;
            }//else
        }//while
        return false;
    }
};

GitHub测试程序源码