Leetcode120 Triangle (DP经典题）

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本文讨论了如何在给定三角形数组中找到从顶点到底端的最小路径和，提供了三种不同的动态规划解法，包括直接DP、滚动数组优化和01背包空间优化技巧。

Triangle
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Given a triangle array, return the minimum path sum from top to bottom.

For each step, you may move to an adjacent number of the row below. More formally, if you are on index i on the current row, you may move to either index i or index i + 1 on the next row.

Example 1:

Input: triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
Output: 11
Explanation: The triangle looks like:
2
3 4
6 5 7
4 1 8 3
The minimum path sum from top to bottom is 2 + 3 + 5 + 1 = 11 (underlined above).
Example 2:

Input: triangle = [[-10]]
Output: -10

Constraints:

1 <= triangle.length <= 200
triangle[0].length == 1
triangle[i].length == triangle[i - 1].length + 1
-104 <= triangle[i][j] <= 104

Follow up: Could you do this using only O(n) extra space, where n is the total number of rows in the triangle?

解法1：最直接的DP。

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int n = triangle.size();
        if (n == 0) return 0;
        
        vector<vector<int>> dp(n, vector<int>(n, 0));
        dp[0][0] = triangle[0][0];
        //line i (start from 0) has i + 1 numbers
        for (int i = 1; i < n; i++) {
            dp[i][0] = dp[i - 1][0] + triangle[i][0];
            dp[i][i] = dp[i - 1][i - 1] + triangle[i][i];
            for (int j = 1; j < i; j++) {
                dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j]) + triangle[i][j];
            }
        }
        int res = dp[n - 1][0];
        for (int i = 1; i < n; i++) {
            if (res > dp[n - 1][i]) res = dp[n - 1][i];
        }
        return res;
    }
};

解法2：滚动数组实现空间优化

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int n = triangle.size();
        if (n == 0) return 0;
        
        vector<vector<int>> dp(2, vector<int>(n, 0));
        dp[0][0] = triangle[0][0];
        //line i (start from 0) has i + 1 numbers
        for (int i = 1; i < n; i++) {
            dp[i % 2][0] = dp[(i - 1) % 2][0] + triangle[i][0];
            dp[i % 2][i] = dp[(i - 1) % 2][i - 1] + triangle[i][i];
            for (int j = 1; j < i; j++) {
                dp[i % 2][j] = min(dp[(i - 1) % 2][j - 1], dp[(i - 1) % 2][j]) + triangle[i][j];
            }
        }
        int res = dp[(n - 1) % 2][0];
        for (int i = 1; i < n; i++) {
            if (res > dp[(n - 1) % 2][i]) res = dp[(n - 1) % 2][i];
        }
        return res;
    }
};

解3：
类似01背包的空间优化。注意j的顺序是从大到小。

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        int n = triangle.size();
        if (n == 0) return 0;
        
        vector<int> dp(n, INT_MIN);
        dp[0] = triangle[0][0];
        //line i (start from 0) has i + 1 numbers
        for (int i = 1; i < n; i++) {
            // j 的处理是从大到小 
            dp[i] = dp[i - 1] + triangle[i][i];           
            for (int j = i - 1; j > 0; j--) {
                dp[j] = min(dp[j - 1], dp[j]) + triangle[i][j];
            }
            dp[0] = dp[0] + triangle[i][0]; 
        }
        int res = dp[0];
        for (int i = 1; i < n; i++) {
            if (res > dp[i]) res = dp[i];
        }
        return res;
    }
};