Archive for November 23rd, 2013
- In: leetcode
- 5 Comments
遇见这种matrix题就腿发软，bf给讲了个四条边bound的方法，立刻15分钟bug free了。高兴够呛。
public int[][] generateMatrix(int n) {
int[][] res = new int[n][n];
int k = 1;
int top = 0, bottom = n - 1, left = 0, right = n - 1;
while (left < right && top < bottom) {
for (int j = left; j < right; j++) {
res[top][j] = k++;
}
for (int i = top; i < bottom; i++) {
res[i][right] = k++;
}
for (int j = right; j > left; j--) {
res[bottom][j] = k++;
}
for (int i = bottom; i > top; i--) {
res[i][left] = k++;
}
left++;
right--;
top++;
bottom--;
}
if (n % 2 != 0)
res[n / 2][n / 2] = k;
return res;
}
Tags: matrix
[leetcode] Valid Number
Posted on: November 23, 2013
- In: leetcode
- 7 Comments
这就是典型的try and fail，尼玛谁知道什么算是数什么不算啊？让我用眼睛看我也不知道啊！总之，试出来的规则是这样的：
- AeB代表A * 10 ^ B
- A可以是小数也可以是整数，可以带正负号
- .35, 00.神马的都算valid小数；就”.”单独一个不算
- B必须是整数，可以带正负号
- 有e的话，A,B就必须同时存在
算法就是按e把字符串split了，前面按A的法则做，后面按B做。
public boolean isNumber(String s) {
s = s.trim();
if (s.length() > 0 && s.charAt(s.length() - 1) == 'e')
return false; //avoid "3e" which is false
String[] t = s.split("e");
if (t.length == 0 || t.length > 2)
return false;
boolean res = valid(t[0], false);
if (t.length > 1)
res = res && valid(t[1], true);
return res;
}
private boolean valid(String s, boolean hasDot) {
if (s.length() > 0 && (s.charAt(0) == '+' || s.charAt(0) == '-')) //avoid "1+", "+", "+."
s = s.substring(1);
char[] arr = s.toCharArray();
if (arr.length == 0 || s.equals("."))
return false;
for (int i = 0; i < arr.length; i++) {
if (arr[i] == '.') {
if (hasDot)
return false;
hasDot = true;
} else if (!('0' <= arr[i] && arr[i] <= '9')) {
return false;
}
}
return true;
}
Tags: math
