Day 4

剑指 offer 11 旋转数组的最小数字

题目描述

解答

//二分查找的时间复杂度是 log n
class Solution {
    public int minArray(int[] numbers) {
        int low=0,high=numbers.length-1,mid;
        while(low<high){
            mid=(low+high)/2;
            if(numbers[mid]<numbers[high]){
                high=mid;
            }else if(numbers[mid]>numbers[high]){
                low+=1;
            }else high-=1;
        }
        return numbers[low];
    }
}

剑指 offer 13 机器人的运动范围

题目描述

解答

class Solution {
    public int movingCount(int m, int n, int k) {
        boolean[][] visited = new boolean[m][n];
        return dfs(0, 0, m, n, k, visited);
    }

    private int dfs(int i, int j, int m, int n, int k, boolean visited[][]) {
        // i表示当前所在行，j表示当前所在列
        // visited[i][j]=true 表明i行j列的元素已被访问过
        if (i < 0 || i >= m || j < 0 || 
        j >= n || (i/10 + i%10 + j/10 + j%10) > k || visited[i][j]) {
            return 0;
        }
        visited[i][j] = true;
        return dfs(i + 1, j, m, n, k, visited) 
               +dfs(i, j + 1, m, n, k, visited)
               +1;
    }
}