Leetcode 题解系列（十二）

120. Triangle

题目要求

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.

题目分析

动态规划

由于到达点[i, j]的路径必须经过点[i - 1, j - 1]或者点[i - 1, j]，假设到达点[i, j]的最小路径为dp(i, j)，经过该点的代价为Ci,j，那么有：

dp(i,j)=⎧⎩⎨⎪⎪⎪⎪⎪⎪Ki,j,dp(i−1,j)+Ki,j,dp(i−1,j−1)+Ki,j,min{dp(i−1,j),dp(i−1,j−1)},i = 0j = 0j = iothers

时间复杂度和空间复杂度为O(N2)
为了方便，可以在每一行的前后各加距离为正无穷的点，简化逻辑。
代码如下：

#define MAX 1000000class Solution {public:    int minimumTotal(vector<vector<int>>& triangle) {        int depth = triangle.size();        if (depth == 0) return 0;        int width = triangle[depth - 1].size();        vector<vector<int>> dp;        dp.push_back({MAX, triangle[0][0], MAX});        for (int i = 1; i < depth; ++i) {            dp.push_back(vector<int>(i + 3, MAX));            for (int j = 1; j <= i + 1; ++j) {                dp[i][j] = std::min(dp[i - 1][j], dp[i - 1][j - 1]) + triangle[i][j - 1];            }        }        int min = 1000000;        for (const auto& i: dp[depth - 1]) {            if (min > i) min = i;        }        return min;    }};

降低空间复杂度

由于每一行只依赖于上一行的结果，所以可以进行覆盖，但要记得保存dp(i−1,j−1)。

#define MAX 1000000class Solution {public:    int minimumTotal(vector<vector<int>>& triangle) {        int depth = triangle.size();        if (depth == 0) return 0;        int width = triangle[depth - 1].size();        vector<int> dp(width + 2, MAX);        dp[1] = triangle[0][0];        int min = MAX;        for (int i = 1; i < depth; ++i) {            int prev = dp[0];            for (int j = 1; j < i + 2; ++j) {                int tmp = dp[j];                dp[j] = std::min(dp[j], prev) + triangle[i][j - 1];                prev = tmp;            }        }        for (const auto& i: dp) {            if (min > i) min = i;        }        return min;    }};