offer-13


地上有一个m行n列的方格,从坐标 [0,0] 到坐标 [m-1,n-1] 。一个机器人从坐标 [0, 0] 的格子开始移动,它每次可以向左、右、上、下移动一格(不能移动到方格外),也不能进入行坐标和列坐标的数位之和大于k的格子。例如,当k为18时,机器人能够进入方格 [35, 37] ,因为3+5+3+7=18。但它不能进入方格 [35, 38],因为3+5+3+8=19。请问该机器人能够到达多少个格子?

示例 1:

输入:m = 2, n = 3, k = 1
输出:3

示例 2:

输入:m = 3, n = 1, k = 0
输出:1

提示:

  • 1 <= n,m <= 100
  • 0 <= k <= 20

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public int movingCount(int m, int n, int k) {
        if (m < 0 || n < 0) {
            return 0;
        }

        Boolean[][] canVisit = new Boolean[m][n];
        visitNext(canVisit, m, n, 0, 0, k);
        int result = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (canVisit[i][j] != null && canVisit[i][j]) {
                    result++;
                }
            }
        }

        return result;
    }

    private void visitNext(Boolean[][] canVisit, int m, int n, int curR, int curC, int k) {
        if (curR < 0 || curR >= m || curC < 0 || curC >= n || canVisit[curR][curC] != null) {
            return;
        }

        int sum = 0;
        int temp = curR;
        while (temp > 0) {
            sum += temp % 10;
            temp /= 10;
        }
        temp = curC;
        while (temp > 0) {
            sum += temp % 10;
            temp /= 10;
        }

        if (sum > k) {
            canVisit[curR][curC] = false;
            return;
        }
        canVisit[curR][curC] = true;
        visitNext(canVisit, m, n, curR - 1, curC, k);
        visitNext(canVisit, m, n, curR + 1, curC, k);
        visitNext(canVisit, m, n, curR, curC - 1, k);
        visitNext(canVisit, m, n, curR, curC + 1, k);
    }
}
//leetcode submit region end(Prohibit modification and deletion)

文章作者: 倪春恩
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