剑指offer编程题——03 二维数组中的查找

// 二维数组中的查找
/*
1 2  8  9
2 4  9 12
4 7 10 13
6 8 11 15
*/

#include <iostream>
#define LEN 4
using namespace std;

bool SearchKey(int a[][LEN], int key){

bool flag = false;
if (a == NULL)
return flag;

int row = 0;
int col = LEN - 1;
while (col >= 0 && row < LEN){
if (a[row][col] == key){
flag = true;
break;
}
else if (a[row][col] == key)
++row;
else{
while (a[row][col] > key && col >= 0)
--col;
}
}
return flag;
//while (a[row][col] < key && row < LEN){
//	++row;
//	sign = 1;
//}
//if (a[row][col] == key)
//	flag = true;
/* 若Key不存在，则会出现无限循环的情况，如何破？
*	此时必然会最终找到左下角的数，行列下标总有一个为0
*/
//if (row == LEN - 1 || col == LEN - 1) 
//	return false;
}

/*
*  The Best Way
1、首先检查形参，确保传进来的数组是有效的
2、不会出现大量true或false，而是用一个变量found记录
*/
bool Find(int *matrix, int rows, int cols, int key){
bool found = false;
if (matrix != NULL && rows > 0 && cols > 0){
int r = 0;
int c = cols - 1;
while (r < rows - 1 && c >= 0){
if (matrix[r*c + c] == key){
found == true;
break;
}
else if (matrix[r*c + c] > key)
--c;
else
++r;
}
}
return found;
}

void main03(){

int a[LEN][LEN];
for (int i = 0; i < LEN; ++i)
for (int j = 0; j < LEN; ++j)
cin>> a[i][j];

cout << endl;

//int a[LEN][LEN] = { { 1, 2, 8, 9 },{ 2, 4, 9, 12 },{ 4, 6, 10, 13 },{6, 8, 11, 15} };

for (int i = 0; i < LEN; ++i){
for (int j = 0; j < LEN; ++j)
cout << a[i][j] << " ";
cout << endl;
}

int key;
cout << endl << "You want to find: ";
cin >> key;

bool flag = SearchKey(a, key);
cout << flag << endl;

cout << endl << endl;
}