[Leetcode] Triangle

Triangle

Given a triangle, find the minimum path sum from top to bottom. Each step of the 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  + < Span style= "color:red" >3  + 5  +  1  = one).

Note:
Bonus Point If-able to does this using only O(n) extra space, where n is the total number of rows in the triangle.

Problem Solving Ideas:

Dynamic planning. Use D to record the shortest path from the root to the page, and I is the current number of rows. Then there are

D[J] = d[j-1] + triangle[i][j],j==i

D[J] = d[0] + triangle[i][0], j==0

D[j] = min (d[j-1], d[j]) + triangle[i][j], others.

Note that the last d[j-1] is used when calculating d[j], so you can press d[i. 0] In order to calculate.

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--) {    //here from high to low                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;}    ;

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[Leetcode] Triangle