LeetCode题解（Java）：120-三角形最小路径和
120. 三角形最小路径和
https://leetcode-cn.com/problems/triangle/
解法一：动态规划（二维）
- DP方程
d p [ i ] [ j ] = m i n ( d p [ i + 1 ] [ j ] , d p [ i + 1 ] [ j + 1 ] ) + t r i a n g l e [ i ] [ j ] dp[i][j] = min(dp[i + 1][j], dp[i + 1][j + 1]) + triangle[i][j] dp[i][j]=min(dp[i+1][j],dp[i+1][j+1])+triangle[i][j]
- 代码
class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
int m = triangle.size();
int[][] dp = new int[m + 1][m + 1];
for (int i = m - 1; i >= 0; i--) {
for (int j = 0; j <= i; j++) {
dp[i][j] = triangle.get(i).get(j) + Math.min(dp[i + 1][j], dp[i + 1][j + 1]);
}
}
return dp[0][0];
}
}
解法二：动态规划（一维）
- DP方程
d p [ j ] = m i n ( d p [ j ] , d p [ j + 1 ] ) + t r i a n g l e [ i ] [ j ] dp[j] = min(dp[j], dp[j + 1]) + triangle[i][j] dp[j]=min(dp[j],dp[j+1])+triangle[i][j]
- 代码
class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
int m = triangle.size();
int[] dp = new int[m];
List<Integer> lastRow = triangle.get(m - 1);
for (int i = 0; i < lastRow.size(); i++) {
dp[i] = lastRow.get(i);
}
for (int i = m - 2; i >= 0; i--) {
for (int j = 0; j <= i; j++) {
dp[j] = triangle.get(i).get(j) + Math.min(dp[j], dp[j + 1]);
}
}
return dp[0];
}
}