65 lines
2.0 KiB
C++
65 lines
2.0 KiB
C++
// Source : https://leetcode.com/problems/minimum-falling-path-sum/
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// Author : Hao Chen
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// Date : 2019-01-30
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/*****************************************************************************************************
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*
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* Given a square array of integers A, we want the minimum sum of a falling path through A.
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*
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* A falling path starts at any element in the first row, and chooses one element from each row. The
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* next row's choice must be in a column that is different from the previous row's column by at most
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* one.
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*
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* Example 1:
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*
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* Input: [[1,2,3],[4,5,6],[7,8,9]]
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* Output: 12
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* Explanation:
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* The possible falling paths are:
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*
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* [1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9]
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* [2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9]
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* [3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]
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*
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* The falling path with the smallest sum is [1,4,7], so the answer is 12.
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*
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* Note:
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*
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* 1 <= A.length == A[0].length <= 100
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* -100 <= A[i][j] <= 100
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******************************************************************************************************/
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class Solution {
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private:
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int min(int x, int y) {
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return x < y ? x: y;
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}
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int min( int x, int y, int z) {
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return min(min(x, y),z);
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}
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public:
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int minFallingPathSum(vector<vector<int>>& A) {
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int m = INT_MAX;
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for (int i=0; i<A.size(); i++) {
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for (int j=0; j<A[i].size(); j++){
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//find the minimal item in previous row, and add it into the current item
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if (i > 0) {
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if (j == 0 ){
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A[i][j] += min( A[i-1][j], A[i-1][j+1]);
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} else if ( j + 1 == A[i].size()) {
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A[i][j] += min( A[i-1][j], A[i-1][j-1]);
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}else {
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A[i][j] += min( A[i-1][j], A[i-1][j-1], A[i-1][j+1]);
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}
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}
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if ( i + 1 == A.size() ) {
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m = min(m, A[i][j]);
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}
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}
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}
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return m;
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}
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};
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