LeetCode 120. Triangle

120. Triangle 
DescriptionHintsSubmissionsSolutions
Total Accepted: 101928
Total Submissions: 306497
Difficulty: Medium
Contributor: LeetCode
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.

public class Solution {    public int minimumTotal(List<List<Integer>> triangle) {        List<Integer> list = null;        int m= triangle.size();        if(m == 1) return triangle.get(0).get(0);        int k = 0;        for(int i = 0; i < m; i++) {            k += triangle.get(i).size();//元素总个数        }        int[] dp = new int[k];        dp[0] = triangle.get(0).get(0);        // if(triangle.get(0).size() == 1) return triangle.get(0).get(0);        int s = 1;        for(int i = 1; i < m; i++) {//            int s = (int) (Math.pow(2,i) - 1);//数组中起始位置            list = triangle.get(i);            int n = list.size();            dp[s] = dp[s-n+1] + list.get(0);            for(int j = 1; j < n-1; j++) {                dp[s+j] = Math.min(dp[s+j-n],dp[s+j-n+1]) + list.get(j);            }            dp[s+n-1] = dp[s-1] + list.get(n-1);//最后一个元素             s += n;        }        int min = 10000;        for(int i = k - 1; i >= k - triangle.get(m-1).size(); i--) {        if(min > dp[i]) min = dp[i];         }        return min;    }}