LeetCode | Minimum Path Sum

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

public class Solution {
	
    //思想与解法同二维dynamic programming的苹果收集问题
	public int minPathSum(int[][] grid) {
		int m = grid.length;    // 取得rows
		int n = grid[0].length; // 取得cols

		// 状态变量:states[i][j]表示达到点(i-1, j-1)的最小路径和
		int states[][] = new int[m][n];

		int sum_1 = 0;
		for (int i = 0; i < n; i++) {   //填充状态变量第0行
			sum_1 += grid[0][i];
			states[0][i] = sum_1;
		}

		int sum_2 = 0;
		for (int j = 0; j < m; j++) {   //填充状态变量第0列
			sum_2 += grid[j][0];
			states[j][0] = sum_2;
		}
		
		if (m < 2 || n < 2) {           //避免下面的数组发生(index = -1)溢出
			return states[m - 1][n - 1];
		}

		// 状态转移方程:达到点(i,j),要么从上面到达,要么从左面达到
		// states[i][j] = min( states[i-1][j]+grid[i][j], states[i][j-1]+grid[i][j] ), { i>=1, j>=1}
		for (int i = 1; i < m; i++) {
			for (int j = 1; j < n; j++) {
				states[i][j] = Math.min(states[i-1][j] + grid[i][j], states[i][j-1] + grid[i][j]);
			}
		}

		return states[m - 1][n - 1];
	}

}



原文地址:https://www.cnblogs.com/dosmile/p/6444447.html