Minimum Path Sum

题目原型：

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.

基本思路：

这是一道动态规划题，从图中可以看出，我们没次从右上角斜向下走来更新第n层的值，借助此数组更新数据，知道最后退出。

	public int minPathSum(int[][] grid)
	{
		int lenOfRow = grid.length;
		int lenOfCol = grid[0].length;
		if (grid == null || lenOfRow == 0)
			return 0;
		
		int row = 0;
		int col = 0;

		int startRow = 0;
		int startCol = 0;
		
		while (true)
		{
			//如果起始点没达到最后一列
			if(startCol+1<lenOfCol)
				startCol+=1;
			//如果起始点没达到最后一行
			else if(startRow+1<lenOfRow)
				startRow+=1;
			//如果达到最后一行和一列，即终点，则退出
			else
				break;
			row = startRow;
			col = startCol;
			//一般情况下，每次保证节点沿着左下方运行
			while(row<lenOfRow&&col>=0)
			{
				//假如在第一行,则此节点的值只能由次行中的前一个节点决定
				if(row==0)
					grid[row][col] = grid[row][col]+grid[row][col-1];
				//如果在最后一列，则次节点的值只能由此列的前一个元素决定
				else if(col==0)
					grid[row][col] = grid[row][col]+grid[row-1][col];
				//否则，由此节点的前一行和前一列共同决定
				else
					grid[row][col] = (grid[row-1][col]<grid[row][col-1]?grid[row-1][col]:grid[row][col-1])+grid[row][col];
				row++;
				col--;
			}
		}

		return grid[lenOfRow-1][lenOfCol-1];
	}

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Minimum Path Sum

原文：http://blog.csdn.net/cow__sky/article/details/21325831