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面试题77:二维数组递减路径

2016-04-12 20:58 393 查看
给定一个二维数组,从左下方元素开始,找出最长一条递减路径,路径只能往右方和上方移动。

例如:

1 3 2

5 4 6

7 9 8

则最长路径为:7->5->4->3->2

思路:

回溯思想。

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

void FindPathCore(int row,int col,vector<vector<int>> num, vector<vector<int>> &re,vector<int>&path)
{
int size1 = num.size();
int size2 = num[0].size();
if (row >= size1 || col >= size2 || row<0 || col<0) return;
path.push_back(num[row][col]);
if (row == 0 && col == size2 - 1) re.push_back(path);
if (col<size2 - 1 && ((row == 0 && num[row][col] > num[row][col + 1]) || (num[row][col] > num[row][col + 1])))
FindPathCore(row, col + 1, num, re, path);
else if (row>0 && ((col == size2 - 1 && num[row][col] > num[row - 1][col]) || (num[row][col] > num[row - 1][col])))
FindPathCore(row - 1, col, num, re, path);
else re.push_back(path);
path.pop_back();
}

int GetLargest(vector<vector<int>> re)
{
if (re.size() == 0) return 0;
unsigned int max = re[0].size();
for (unsigned int i = 1; i < re.size(); i++)
{
if (re[i].size() > max) max = re[i].size();
}
return max;
}

int FindPath(vector<vector<int>> num)
{
int size1 = num.size();
if (size1 == 0) return 0;
int size2 = num[0].size();
vector<vector<int>> re;
vector<int> path;
for (int i = size1-1; i >= 0; i--)
{
for (int j = 0; j < size2; j++)
{
if (i == 0 && j == 0) FindPathCore(i, j, num, re, path);
else if (i == 0 && num[i][j] >= num[i][j - 1]) FindPathCore(i, j, num, re, path);
else if (j == 0 && num[i][j] >= num[i - 1][j]) FindPathCore(i, j, num, re, path);
else if (num[i][j] >= num[i][j - 1] && num[i][j] >= num[i - 1][j]) FindPathCore(i, j, num, re, path);
}
}
return GetLargest(re);
}

int main(){
int N = 5;
vector<vector<int>> num(N, vector<int>(N, 0));
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
num[i][j] = i + j;
}
}
num[2][1] = 1;
num[2][2] = 0;
num[2][3] = -1;
num[2][4] = -2;
num[1][4] = -3;
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
{
cout << num[i][j] << " ";
}
cout << endl;
}
cout << FindPath(num) << endl;
return 0;
}
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