Leetcode 120. 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.

s思路：

1. 最小路径和，从上往下，下面的数的坐标只能是上一个数的坐标或坐标加1。枚举所有路径，然后找出最小值？

2. 枚举代码功能正确，但是TLE?那必须是大量重复计算所致。分析如下：



如上图，2到5的路径有多条，也就是说从5开始往下遍历需要遍历多次，这就重复计算了！完全可以从上往下计算到每个位置的最小和，然后不断迭代进行，也就是在每个位置计算最小和保存起来。这就是DP的思路！

3. 以后做recursive的题，先看看是否会有重复计算，如有，则用DP！！

//方法1：recursive的枚举所有路径。TLE?
class Solution {
public:

void helper(vector<vector<int>>& triangle,int level,int pos,int&path,int cur){

if(level==triangle.size()){
path=path<cur?path:cur;
return;
}

helper(triangle,level+1,pos,path,cur+triangle[level][pos]);
if(pos+1<triangle[level].size())
helper(triangle,level+1,pos+1,path,cur+triangle[level][pos+1]);

}
int minimumTotal(vector<vector<int>>& triangle) {
//
int path=INT_MAX;
helper(triangle,0,0,path,0);
return path;
}
};

//方法2：dp
class Solution {
public:

int minimumTotal(vector<vector<int>>& triangle) {
//
int n=triangle.size();
int path=INT_MAX;

for(int i=1;i<n;i++){
triangle[i][0]+=triangle[i-1][0];
triangle[i][i]+=triangle[i-1][i-1];
for(int j=1;j<i;j++){
triangle[i][j]+=min(triangle[i-1][j],triangle[i-1][j-1]);
}
}
for(int i=0;i<n;i++){
path=min(path,triangle[n-1][i]);
}
return path;
}
};