Triangle
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[ [2], [3,4], [6,5,7], [4,1,8,3] ]
The minimum path sum from top to bottom is 11
(i.e., 2 + 3 + 5 + 1 =
11).
Note:
Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle.
此题可谓一波三折, 先贴一个本人的错误代码:
class Solution { public: int minimumTotal(vector<vector<int> > &triangle) { int row = triangle.size(); if(row==1) return triangle[0][0]; int result[row]; memset(result,0,sizeof(result)); int i,j; int temp1,temp2; result[0] = result[1] = triangle[0][0]; for(i=1;i<row;i++) { temp1 = result[0]; temp2 = result[i-1]; for(j=1;j<i;j++) { if(result[j-1]<result[j]) result[j] = result[j-1]+triangle[i][j]; else result[j] = result[j]+triangle[i][j]; } result[0] = temp1 + triangle[i][0]; result[i] = temp2 + triangle[i][i]; } sort(result,result+row); return result[0]; } };错误原因是, 在更新result[j]时,会对下一次的判断有影响. 因为下一次要用到restult[j-1]
正确的代码只是稍微修改了一下, 从后往前更新, 这样就避免了影响
class Solution { public: int minimumTotal(vector<vector<int> > &triangle) { int row = triangle.size(); if(row==1) return triangle[0][0]; int result[row]; memset(result,0,sizeof(result)); int i,j; int temp1,temp2; result[0] = result[1] = triangle[0][0]; for(i=1;i<row;i++) { temp1 = result[0]; temp2 = result[i-1]; for(j=i-1;j>=1;j--) { if(result[j-1]<result[j]) result[j] = result[j-1]+triangle[i][j]; else result[j] = result[j]+triangle[i][j]; } result[0] = temp1 + triangle[i][0]; result[i] = temp2 + triangle[i][i]; } sort(result,result+row); return result[0]; } };
每天早上叫醒你的不是闹钟,而是心中的梦~