12
08/2015
[LeetCode] 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.
解题思路：
动态规划。用d来记录从根到页的最短路径，i为当前行数。则有
d[j] = d[j-1] + triangle[i][j]，j==i时
d[j] = d[0] + triangle[i][0], j==0时
d[j] = min(d[j-1], d[j]) + triangle[i][j]，其他。
注意计算d[j]时，会用到上一次的d[j-1]，因此可以按d[i..0]的顺序来计算。
class Solution {
public:
int minimumTotal(vector<vector<int>>& triangle) {
int n = triangle.size();
if(n==0){
return 0;
}
if(n==1){
return triangle[0][0];
}
int d[n];
d[0] = triangle[0][0];
for(int i=1; i<n; i++){
for(int j=i; j>=0; j--){ //这里从由高到低
if(j==i){
d[j] = d[j-1] + triangle[i][j];
}else if(j==0){
d[j] = d[0] + triangle[i][0];
}else{
d[j] = min(d[j-1], d[j]) + triangle[i][j];
}
}
}
int minPath = d[0];
for(int i=1; i<n; i++){
minPath = min(d[i], minPath);
}
return minPath;
}
};转载请注明：康瑞部落 » [LeetCode] Triangle
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