题目

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

  1. [
  2.      [2],
  3.     [3,4],
  4.    [6,5,7],
  5.   [4,1,8,3]
  6. ]

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.

题意

给定一个三角形矩阵，求出从上到下的最小路径之和。

思路

动态规划：dp[i][j]代表从上到下走到(i, j)时的最小路径和，很容易看出：

0120. Triangle (M) - 图1%0A#card=math&code=dp%5Bi%5D%5Bj%5D%3Dtriangle%5Bi%5D%5Bj%5D%2Bmin%28dp%5Bi-1%5D%5Bj-1%5D%2C%5C%20dp%5Bi-1%5D%5Bj%5D%29%0A&id=tk46s)

对于空间复杂度0120. Triangle (M) - 图2#card=math&code=O%28N%29&id=CyYwx)的要求，可以使用滚动数组进行优化，这时候需要从最底层往最高层走，同样有公式：

0120. Triangle (M) - 图3%0A#card=math&code=dp%5Bj%5D%3Dtriangle%5Bi%5D%5Bj%5D%2Bmin%28dp%5Bj%5D%2C%5C%20dp%5Bj%2B1%5D%29%0A&id=J0g8V)

(实际操作中未使用滚动数组优化的动态规划也可以从下往上走，这样走到顶时得到的和就是最小路径和。)

代码实现

Java

动态规划

  1. class Solution {
  2.     public int minimumTotal(List<List<Integer>> triangle) {
  3.         int[][] dp = new int[triangle.size()][triangle.size()];
  5.         for (int i = triangle.size() - 1; i >= 0; i--) {
  6.             for (int j = 0; j < triangle.get(i).size(); j++) {
  7.                 int curNum = triangle.get(i).get(j);
  9.                 if (i == triangle.size() - 1) {
  10.                     dp[i][j] = curNum;
  11.                     continue;
  12.                 }
  14.                 dp[i][j] = curNum + Math.min(dp[i + 1][j], dp[i + 1][j + 1]);
  15.             }
  16.         }
  18.         return dp[0][0];
  19.     }
  20. }

滚动数组优化

  1. class Solution {
  2.     public int minimumTotal(List<List<Integer>> triangle) {
  3.         int[] dp = new int[triangle.size()];
  5.         for (int i = triangle.size() - 1; i >= 0; i--) {
  6.             for (int j = 0; j < triangle.get(i).size(); j++) {
  7.                 int curNum = triangle.get(i).get(j);
  9.                 if (i == triangle.size() - 1) {
  10.                     dp[j] = curNum;
  11.                     continue;
  12.                 }
  14.                 dp[j] = curNum + Math.min(dp[j], dp[j + 1]);
  15.             }
  16.         }
  18.         return dp[0];
  19.     }
  20. }

JavaScript

  1. /**
  2.  * @param {number[][]} triangle
  3.  * @return {number}
  4.  */
  5. var minimumTotal = function (triangle) {
  6.   let ans = Number.MAX_SAFE_INTEGER
  7.   const n = triangle.length
  8.   const dp = new Array(n).fill(0)
  10.   for (let i = 0; i < n; i++) {
  11.     for (let j = i; j >= 0; j--) {
  12.       if (j === 0) {
  13.         dp[j] = triangle[i][j] + dp[j]
  14.       } else if (j === i) {
  15.         dp[j] = triangle[i][j] + dp[j - 1]
  16.       } else {
  17.         dp[j] = triangle[i][j] + Math.min(dp[j], dp[j - 1])
  18.       }
  20.       if (i === n - 1) ans = Math.min(ans, dp[j])
  21.     }
  22.   }
  24.   return ans
  25. }