[LeetCode 62] Unique Paths(教科書般的動態規劃)
題目內容
62 Unique Paths
A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below).
The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below).
How many possible unique paths are there?
Above is a 3 x 7 grid. How many possible unique paths are there?
Note: m and n will be at most 100.
題目來源
題目簡述
計算機器人從起始點到達終點不同路徑的總數
題目分析
簡單的組合數計算因為數字溢位顯然不可以。
我們假設起點為(0,0),終點為(m,n)。設到達(x,y)點的路徑總數為A(x,y)。顯然A(x,y)=A(x-1,y)+A(x,y-1)。根據這個狀態轉移方程從頭計算即可得到A(m,n)。
但是我們可以使用一維陣列B[x]記錄一行行內每個座標資料,並對每行迴圈更新,大大減少空間複雜度。利用上式對B[x]進行計算時,注意到等號右邊A(x-1,y)就是本次迴圈中剛剛的更新的B[x-1],A(x,y-1)就是上次迴圈更新的B[x],本次迴圈該資料尚未更新。整理得B[x]=B[x]+B[x-1]。外迴圈進行n-1次後得到B[m]就是起點到終點的路徑總數。
程式碼示例
class Solution {
public:
int uniquePaths(int m, int n) {
vector<int> v(m,1);
for(int i=1;i!=n;i++)
{
for(int j=1;j!=m;j++)
v[j]+=v[j-1];
}
return v[m-1];
}
};
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