编程题1

题目描述：寻找一条从左上角arr[0][0]到右下角arr[m-1][n-1]的路线，使得沿途经过的数组中的整数之和最小

代码：

java版实现：

public class FindRoute {

    public static int getMinPath(int[][] arr){
        if (arr == null || arr.length == 0){
            return 0;
        }
        int row = arr.length;
        int col = arr[0].length;

        // 用来保存计算的中间值
        int[][] cache = new int[row][col];
        cache[0][0] = arr[0][0];

        for (int i = 1; i < col; i++){
            cache[0][i] = cache[0][i - 1] + arr[0][i];
        }
        for (int j = 1; j < row; j++){
            cache[j][0] = cache[j - 1][0] + arr[j][0];
        }
        System.out.println("["+(0)+","+0+"]:  " + arr[0][0]);
        // 在遍历二维数组的过程中不断把计算结果保存到cache中
        for (int i = 1; i < row; i++){
            for (int j = 1; j < col; j++){
                if (cache[i - 1][j] > cache[i][j - 1]){ // 可以确定选择的路线为arr[i][j - 1]
                    cache[i][j] = cache[i][j - 1] + arr[i][j];
                    System.out.println("["+i+","+(j - 1)+"]:  " + arr[i][j - 1]);
                }else {   // 可以确定选择的路线为arr[i - 1][j]
                    cache[i][j] = cache[i - 1][j] + arr[i][j];
                    System.out.println("["+(i - 1)+","+j+"]:  " + arr[i - 1][j]);
                }
            }
        }
        System.out.println("[" +(row - 1) + "," + (col - 1) + "]" + arr[row - 1][col - 1]);
        return cache[row - 1][col - 1];
    }


    public static void main(String[] args){
        int[][] arr = {

                {1,4,3,3},
                {8,7,5,5},
                {2,1,5,5},
                {2,1,5,5}
        };
        System.out.println("路径：");
        System.out.println("最小值为："+getMinPath(arr));
    }
}