Lexi's Leetcode solutions

Archive for November 2nd, 2013

这题完全没印象,后来发现是之前做用了string.indexOf这个函数cheat成功了。这也是two pointers一快一慢的练习题,挺难做的,需要重做。

  • i:当前unique substring的头
  • j:当前unique substring的尾
  • 注意最后可能忘记算最后一段的长度了(因为dup那个条件已经不存在)
public int lengthOfLongestSubstring(String s) {
    int i = 0, len = 0, n = s.length();
    char A[] = s.toCharArray();
    Set<Character> set = new HashSet<Character>();
    for (int j = 0; j < n; j++) {
        if (set.contains(A[j])) { //found dup
            len = Math.max(len, j - i);
            while (i < n && A[i] != A[j]) {
                set.remove(A[i]);
                i++;
            }
            i++;
        } else {
            set.add(A[j]);
        }
    }
    return Math.max(len, n - i);
}
  • d[i][j]表示从(0, 0)到(i, j)有几种走法,因为只能从上面来或从左边来,所以每次计算只需要d[i – 1][j]和d[i, j – 1],左上角的全都浪费了。
  • f[i]表示从(0,0)到(当前row, j)有几种走法,上面来的就是自己,左边来的已经算好了,所以 f[j] = f[j] + f[j – 1];
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
    int m = obstacleGrid.length, n = obstacleGrid[0].length;
    // one way DP, f[i] means from (0, 0) to (currRow, i) has how many ways
    int f[] = new int[n];
    f[0] = obstacleGrid[0][0] > 0 ? 0 : 1;
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) { //comes from above or left 
            if (obstacleGrid[i][j] > 0) {
                f[j] = 0;
            } else {
                if (j == 0) //first col only from above
                    f[j] = f[j];
                else 
                    f[j] = f[j] + f[j - 1];
            }
        }
    }
    return f[n - 1];
}
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