LeetCode Devide & Conquer || Search a 2D Matrix II

题目描述

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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
For example,
Consider the following 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]
]
Given target = 5, return true.
Given target = 20, return false.
Difficulty: medium

解题思路

这道题的要求是找出二维数组中是否有某一元素。
如果数组不大，直接将其遍历即可。

class Solution {public:    bool searchMatrix(vector<vector<int> >& matrix, int target) {        int rsize = matrix.size();        if(rsize == 0) return false;        int csize = matrix[0].size();        for(int i = 0; i < rsize; i++){            for(int j = 0; j < csize; j++){                if(matrix[i][j] == target) return true;            }        }        return false;    }};

有一个复杂度更小的方法，直接从二维数组的右上角开始比较，先从每排的最后一个元素开始比较，如果target比最后一个元素大，那么target显然不在这一排。如果target更小，则有可能在这一排。

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

如果按照分治思想，可以先通过每一行的首尾确定target所在的行数，再对该行进行二分查找即可。

class Solution {public:    bool bsearchRow(vector<int>& row,  int target, int top, int bottom){        if(top > bottom) return false;        int mid = bottom + (bottom - top) / 2;        if(target == row[mid]) return true;        else if(target > row[mid]) return bsearchRow(row, target, mid + 1, bottom);        else if(target < row[mid]) return bsearchRow(row, target, top, mid - 1);    }    bool searchMatrix(vector<vector<int> >& matrix, int target) {        int rsize = matrix.size();        if(rsize == 0) return false;        int csize = matrix[0].size();        int row;        int i;        for(i = 0; i < rsize; i++){            if(target >= matrix[i].front() && target <= matrix[i].back()){                row = i;                break;            }        }        if(i == rsize) return false;        else return bsearchRow(matrix[row], target, 0, csize);    }};