1.剑指Offer编程题之二维数组中的查找

题目描述:
在一个二维数组中，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.

解题思路1：
- 从左下角元素开始往上查找，右边元素比这个元素大，上边元素比这个元素小，于是，target比这个元素小就往上找，比这个元素大就往右找.如果出现边界，则查找失败.

C/C++

class Solution{
    public:
    //二维向量这里最外的<>要有空格。否则在比较旧的编译器下无法通过
    bool Find(int target, vector< vector<int> >array){
        if(array.empty()) return false;
        int rows = array.size();
        int cols = array[0].size();
        int i=rows-1,j=0; //左下标元素坐标
        while(i>=0&&j<cols){
            if(target<array[i][j]) --i;
            else if(target>array[i][j]) ++j;
            else return true;
        }
        return false;
    }
};

Python2

class Solution:
    def Find(self,target,array):
        rows = len(array)
        cols = len(array[0])
        i = rows-1
        j = 0
        while j<cols and i>=0:
            if target<array[i][j]:
                i -= 1
            elif target>array[i][j]:
                j += 1
            else:
                return True
        return False

Java

public class Solution{
    public boolean Find(int target, int [][] array){
        int rows = array.length;
        int cols = array[0].length;
        int i = rows-1,j=0;
        while(i>=0&&j<cols){
            if(target<array[i][j])--i;
            else if(target>array[i][j])++j;
            else return true;
        }
        return false;
    }
}

解题思路2：
从右上角开始查找，左边元素比这个元素小，下边元素比这个元素大，于是target比这个元素大，则往下找，比这个元素小则往左找，出现边界查找失败.

C/C++

class Solution{
    public:
    //二维向量这里最外的<>要有空格。否则在比较旧的编译器下无法通过
    bool Find(int target, vector< vector<int> >array){
        if(array.empty())return false;
        int rows = array.size();
        int cols = array[0].size();
        int i=0,j=cols-1;
        while(i<=rows-1&&j>=0){
            if (target<array[i][j])--j;
            else if(target>array[i][j])++i;
            else return true;
        }
        return false;
    }
};

Python2

class Solution:
    def Find(self,target,array):
        rows = len(array)
        cols = len(array[0])
        i = 0
        j = cols-1
        while i<rows-1 and j>=0:
            if target<array[i][j]:
                j -= 1
            elif target>array[i][j]:
                i += 1
            else:
                return True
        return False

Java

public class Solution{
    public boolean Find(int target, int [][] array){
        int rows = array.length;
        int cols = array[0].length;
        int i=0,j=cols-1;
        while(i<=rows-1&&j>=0){
            if(target<array[i][j])--j;
            else if(target>array[i][j])++i;
            else return true;
        }
        return false;
    }
}

解题思路3：
把每一行看成是一个有序递增数组，利用二分法查找，通过遍历每一行得到答案，时间复杂度nlogn.

C/C++

class Solution{
public:
    bool Find(int target,vector< vector<int> >array){
        if(array.empty()) return false;
        int length=array.size();
        for(int i=0;i<length;++i){
            int low = 0;
            int high = array[i].size()-1;
            while(low<=high){
                int mid = (low+high)/2;
                if(target>array[i][mid]) low=mid+1;
                else if(target<array[i][mid]) high=mid-1;
                else return true;
            }
        }
        return false;
    }
};

Python2

class Solution:
    def Find(self,target,array):
        length=len(array)
        for i in range(length):
            low = 0
            high = len(array[i])-1
            while low<=high:
                mid = (low+high)/2
                if target>array[i][mid]:
                    low = mid + 1
                elif target<array[i][mid]:
                    high = mid-1
                else:
                    return True
        return False   

Jave

public class Solution{
    public boolean Find(int target, int [][] array){
        int length=array.length;
        for(int i=0;i<length;++i){
            int low = 0;
            int high = array[i].length-1;
            while(low<=high){
                int mid = (low+high)/2;
                if(target>array[i][mid])low=mid+1;
                else if(target<array[i][mid])high=mid-1;
                else return true;
            }
        }
        return false;
    }
}

本实例均通过牛客网在线编译测试。