leetcode 120. 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).

方法一

bottom up 从下往上

public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
if (triangle == null || triangle.size() == 0) {
return 0;
}
int n = triangle.size();
int[][]  dp = new int

;

for (int i = 0; i < n; i++) {
dp[n - 1][i] = triangle.get(n-1).get(i);
}
for (int i = n - 2; i >= 0; i--) {
for (int j = 0; j < triangle.get(i).size(); j++) {
dp[i][j] = Math.min(dp[i + 1][j], dp[i + 1][j + 1]) + triangle.get(i).get(j);
}
}
return dp[0][0];

}
}

第二种方法，从上往下

public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
if (triangle == null || triangle.size() == 0) {
return 0;
}

int n = triangle.size();
int[][] dp = new int

;
dp[0][0] = triangle.get(0).get(0);

for (int i = 1; i < n; i++) {
dp[i][0] = dp[i - 1][0] + triangle.get(i).get(0);
dp[i][i] = dp[i - 1][i - 1] + triangle.get(i).get(i);
}

for (int i = 1; i < n; i++) {
for (int j = 1; j < i; j++) {
dp[i][j] = Math.min(dp[i - 1][j - 1], dp[i - 1][j]) + triangle.get(i).get(j);
}
}
int min = dp[n - 1][0];
for (int i = 1; i < n; i++) {
min = Math.min(min, dp[n - 1][i]);
}

return min;
}
}