[LeetCode-120] Triangle
Triangle
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.
DP特训第二弹,感觉这题以前做过,就是第N行依赖第N-1行结果,记录上一行及当前行结果就ok~
1 class Solution { 2 public: 3 int minimumTotal(vector<vector<int> > &triangle) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 int row_cnt = triangle.size(); 7 vector<int> min_results1(row_cnt, INT_MAX); 8 vector<int> min_results2(row_cnt, INT_MAX); 9 vector<int>* last_level_result = &min_results1; 10 vector<int>* cur_level_result = &min_results2; 11 vector<int>* swap_result = NULL; 12 min_results1[0] = triangle[0][0]; 13 min_results1[0] = triangle[0][0]; 14 int result = INT_MAX; 15 16 for (int i = 1; i < row_cnt; ++i) { 17 for (int j = 0; j <= i; ++j) { 18 if (0 == j) { 19 (*cur_level_result)[j] = (*last_level_result)[j] + triangle[i][j]; 20 } else if (i == j) { 21 (*cur_level_result)[j] = (*last_level_result)[j - 1] + triangle[i][j]; 22 } else { 23 (*cur_level_result)[j] = min<int>((*last_level_result)[j - 1], (*last_level_result)[j]) + 24 triangle[i][j]; 25 } 26 27 } 28 swap_result = last_level_result; 29 last_level_result = cur_level_result; 30 cur_level_result = swap_result; 31 } 32 for (int i = 0; i < row_cnt; ++i) { 33 if ((*last_level_result)[i] < result) { 34 result = (*last_level_result)[i]; 35 } 36 } 37 return result; 38 } 39 };