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3周前 (04-21) 7次浏览

Triangle (M)

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

``````[
[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.

思路

[dp[i][j]=triangle[i][j]+min(dp[i-1][j-1], dp[i-1][j])
]

[dp[j]=triangle[i][j]+min(dp[j], dp[j+1])
]

(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

``````class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
int[][] dp = new int[triangle.size()][triangle.size()];

for (int i = triangle.size() - 1; i >= 0; i--) {
for (int j = 0; j < triangle.get(i).size(); j++) {
int curNum = triangle.get(i).get(j);

if (i == triangle.size() - 1) {
dp[i][j] = curNum;
continue;
}

dp[i][j] = curNum + Math.min(dp[i + 1][j], dp[i + 1][j + 1]);
}
}

return dp[0][0];
}
}
``````

滚动数组优化

``````class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
int[] dp = new int[triangle.size()];

for (int i = triangle.size() - 1; i >= 0; i--) {
for (int j = 0; j < triangle.get(i).size(); j++) {
int curNum = triangle.get(i).get(j);

if (i == triangle.size() - 1) {
dp[j] = curNum;
continue;
}

dp[j] = curNum + Math.min(dp[j], dp[j + 1]);
}
}

return dp[0];
}
}
``````

JavaScript

``````/**
* @param {number[][]} triangle
* @return {number}
*/
var minimumTotal = function (triangle) {
let ans = Number.MAX_SAFE_INTEGER
const n = triangle.length
const dp = new Array(n).fill(0)

for (let i = 0; i < n; i++) {
for (let j = i; j >= 0; j--) {
if (j === 0) {
dp[j] = triangle[i][j] + dp[j]
} else if (j === i) {
dp[j] = triangle[i][j] + dp[j - 1]
} else {
dp[j] = triangle[i][j] + Math.min(dp[j], dp[j - 1])
}

if (i === n - 1) ans = Math.min(ans, dp[j])
}
}

return ans
}
``````