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.
Runtime: 444 ms
动态规划思想： 从上到下一行一行的扫描并加入总体，dp[i]记录每扫描完一行，并途过本行i元素的最短路径，故dp[]记录到目前为止所有路径的长度。O(n) space
public class Solution {
public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
int[] dp=new int[triangle.size()];
if(triangle.size()==0) return 0;
if(triangle.size()==1) return triangle.get(0).get(0);
dp[0]=triangle.get(0).get(0);
for(int i=1;i<triangle.size();i++){
for(int j=i;j>=0;j--){
if(j==0) dp[0]+=triangle.get(i).get(0);
else if(j<i) dp[j]=triangle.get(i).get(j)+ Math.min(dp[j],dp[j-1]);
else dp[j]=dp[j-1]+triangle.get(i).get(j);
}
}
int ret=Integer.MAX_VALUE;
for(int i=0;i<dp.length;i++){
if(dp[i]<ret) ret=dp[i];
}
return ret;
}
}