120. Triangle(Array; DP)

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.

思路：与119题类似，只是现在DP中存储的不是数组的值，而是到目前为止的minimum sum。赋值DP时为了避免改变上一行结果，也是得从右往左

class Solution {
public:
    int minimumTotal(vector<vector<int>>& triangle) {
        if(triangle.empty()) return 0;
        
        int minValue = INT_MAX;
        vector<int> dp(triangle.size());
        dp[0]=triangle[0][0];
        for(int i = 1; i < dp.size(); i++){
            dp[i] = dp[i-1]+triangle[i][i];
            for(int j = i-1; j > 0; j--){
                dp[j] = min(dp[j-1],dp[j])+triangle[i][j];
            }
            dp[0] += triangle[i][0];
        }
        
        for(int i = 0; i < dp.size(); i++){
            if(dp[i] < minValue){
                minValue = dp[i];
            }
        }
        
        return minValue;
    }
};