题目描述
地上有一个m行和n列的方格。一个机器人从坐标0,0的格子开始移动,每一次只能向左,右,上,下四个方向移动一格,但是不能进入行坐标和列坐标的数位之和大于k的格子。 例如,当k为18时,机器人能够进入方格(35,37),因为3+5+3+7 = 18。但是,它不能进入方格(35,38),因为3+5+3+8 = 19。请问该机器人能够达到多少个格子?
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/**
@author zhengyanan
@date 2017/3/14 @time 17:09
version_3:回溯法
核心思路:
1.从(0,0)开始走,每成功走一步标记当前位置为true,然后从当前位置往四个方向探索,
返回1 + 4 个方向的探索值之和。
2.探索时,判断当前节点是否可达的标准为:
1)当前节点在矩阵内;
2)当前节点未被访问过;
3)当前节点满足limit限制。
3.
运行时间:31ms
占用内存:550k
*/
public int movingCount(int threshold, int rows, int cols) {
boolean[][] visited = new boolean[rows][cols];
return countingSteps(threshold,rows,cols,0,0,visited);
}
public int countingSteps(int limit,int rows,int cols,int r,int c,boolean[][] visited){
if (r < 0 || r >= rows || c < 0 || c >= cols
|| visited[r][c] || bitSum(r) + bitSum(c) > limit) return 0;
visited[r][c] = true;
return countingSteps(limit,rows,cols,r - 1,c,visited)
+ countingSteps(limit,rows,cols,r,c - 1,visited)
+ countingSteps(limit,rows,cols,r + 1,c,visited)
+ countingSteps(limit,rows,cols,r,c + 1,visited)
+ 1;
}
public int bitSum(int t){
int count = 0;
while (t != 0){
count += t % 10;
t /= 10;
}
return count;
}
###############
链接:https://www.nowcoder.com/questionTerminal/6e5207314b5241fb83f2329e89fdecc8
来源:牛客网
class Solution {
private:
int m_nRows;
int m_nCols;
int m_nThreshold;
public:
int movingCount(int threshold, int rows, int cols)
{
int count = 0;
bool * bVisited = new bool[rows*cols];
memset(bVisited, false,rows*cols);
m_nRows = rows;
m_nCols = cols;
m_nThreshold = threshold;
movingCountHelper(bVisited,0,0,count);
delete [] bVisited;
return count;
}
private:
void movingCountHelper(bool * bVisited,int x,int y,int & count){
if (x<0||x>=m_nRows||y<0||y>=m_nCols)
return ;
if (bVisited[getPosition(x,y)]){
return ;
}
if (canBeVisited(x,y)){
bVisited[getPosition(x,y)] = true;
count++;
movingCountHelper(bVisited,x-1,y,count);
movingCountHelper(bVisited,x+1,y,count);
movingCountHelper(bVisited,x,y-1,count);
movingCountHelper(bVisited,x,y+1,count);
}
}
private :
bool canBeVisited(int x,int y){
int count = 0;
while (x!=0){
count+=(x%10);
x/=10;
}
while (y!=0){
count+=(y%10);
y/=10;
}
return count<=m_nThreshold;
}
inline int getPosition(int x,int y){
return x*(m_nCols)+y;
}
};