【leetcode】Triangle
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Question:
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] ]
Anwser 1:
class Solution { public: int minimumTotal(vector<vector<int> > &triangle) { // Start typing your C/C++ solution below // DO NOT write int main() function int rows = triangle.size(); if(0 == rows) return 0; int *minSums = new int[rows]; int *temp = new int[rows]; for(int r = 0; r < rows; r++) { vector<int> vec = triangle[r]; temp[0] = vec[0] + (r > 0 ? minSums[0] : 0); for(int i = 1; i < r; i++) { temp[i] = vec[i] + min(minSums[i-1], minSums[i]); } if(r > 0) { temp[r] = vec[r] + minSums[r-1]; } int *tswap = temp; temp = minSums; minSums = tswap; } int m = minSums[0]; for(int i = 1; i < rows; i++) { if(minSums[i] < m){ m = minSums[i]; } } delete temp; delete minSums; return m; } };
Anwser 2:
class Solution { public: int minimumTotal(vector<vector<int> > &triangle) { // Start typing your C/C++ solution below // DO NOT write int main() function int line = triangle.size(); for(int i = line -2 ; i >= 0; i--) { for(int j = 0; j < triangle[i].size(); j++) { triangle[i][j] += min(triangle[i+1][j], triangle[i+1][j+1]); } } return triangle[0][0]; } };
Anwser 3:
class Solution { public: void run(vector<vector<int> > &triangle, int row, int idx, int curSum, int &minPath) { if (row == triangle.size()) { minPath = min(minPath, curSum); return; } run(triangle, row + 1, idx, curSum + triangle[row][idx], minPath); run(triangle, row + 1, idx + 1, curSum + triangle[row][idx], minPath); } int minimumTotal(vector<vector<int> > &triangle) { // Start typing your C/C++ solution below // DO NOT write int main() function int minPath = INT_MAX; if (triangle.size() == 0) { return 0; } run(triangle, 0, 0, 0, minPath); return minPath; } };
注意点:
1) Judge Small is ok, but Judge Large is error
2) 如果迭代很深的话,容易造成压栈占用内存很高,超出段后会溢出
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