Triangle 三角形求最小路径和 @LeetCode

从底向上DP

package Level3;

import java.util.ArrayList;

/**
* 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.
*
*/
public class S120 {

public static void main(String[] args) {

}

public int minimumTotal(ArrayList<ArrayList<Integer>> triangle) {
int rowLen = triangle.size();

// dp数组用来存储每一格子的最优解
int[][] sum = new int[rowLen][rowLen];

// 最底下一行
ArrayList<Integer> last = triangle.get(triangle.size()-1);
for(int i=0; i<last.size(); i++){
sum[rowLen-1][i] = last.get(i);
}

// Bottom-up DP
for(int i=rowLen-2; i>=0; i--){
ArrayList<Integer> row = triangle.get(i);
for(int j=0; j<=i; j++){
sum[i][j] = Math.min(sum[i+1][j], sum[i+1][j+1]) + row.get(j);
}
}

return sum[0][0];
}
}

public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
for(int i=triangle.size()-2; i>=0; i--) {
List<Integer> list = triangle.get(i);
for(int j=0; j<list.size(); j++) {
int val = list.get(j);
list.set(j, val+Math.min(triangle.get(i+1).get(j), triangle.get(i+1).get(j+1)));
}
}

return triangle.get(0).get(0);
}
}