标签：style blog http color io strong for ar div
LeetCode: 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.
地址：https://oj.leetcode.com/problems/triangle/
算法：开两个大小为n的数组，假设我们已经完成了前面i行的搜索，其中以第i行第j个元素为终节点的最小路径的和存储在其中某个数组里，假设这个数组用指针pre指向。接下去，我们要完成第i+1行的搜索，其中以第i+1行第j个元素为终节点的路径必定是从第i行第j个节点或者第i行第j+1个节点下来的，所有在计算最小路径时只需考虑这两个路径，并把结果存储在另一个数组里，这个数组用指针p指向。由于在每一次搜索的过程中只需要上一次的结果，所以我们只用了两个数组，并且用指针pre和指针p来重复利用这两个数组，这样就可以达到O(n)的空间要求。代码：
1 class Solution {
2 public:
3 int minimumTotal(vector<vector<int> > &triangle) {
4 int n = triangle.size();
5 if(n < 1) return 0;
6 if(n == 1) return triangle[0][0];
7 vector<int> min1(n);
8 vector<int> min2(n);
9 min1[0] = triangle[0][0];
10 vector<int> *pre = &min1;
11 vector<int> *p = &min2;
12 for(int i = 1; i < n; ++i){
13 for(int j = 0; j <= i; ++j){
14 if(j == 0){
15 (*p)[j] = (*pre)[j] + triangle[i][j];
16 }else if(j == i){
17 (*p)[j] = (*pre)[j-1] + triangle[i][j];
18 }else{
19 (*p)[j] = ( (*pre)[j] > (*pre)[j-1] ? (*pre)[j-1] : (*pre)[j] ) + triangle[i][j];
20 }
21 }
22 vector<int> *t = pre;
23 pre = p;
24 p = t;
25 }
26 int result = (*pre)[0];
27 for(int i = 1; i < n; ++i){
28 if (result > (*pre)[i]) result = (*pre)[i];
29 }
30 return result;
31 }
32
33 };
标签：style blog http color io strong for ar div
原文地址：http://www.cnblogs.com/boostable/p/leetcode_triangle.html
