LeetCode118 LeetCode119 LeetCode151 LeetCode202 Java

static public List<List<Integer>> generate(int numRows) {

List<List<Integer>> res = new ArrayList<List<Integer>>();
List<Integer> smallRes = null;

smallRes = new ArrayList<Integer>();

// 大循环内填充List<List<Integer>> res
for (int i = 1; i <= numRows; i++) {
// 小循环内填充List<Integer> smallRes
smallRes = new ArrayList<Integer>();
for (int j = 0; j < i; j++) {
// 取出来res内上一级的内容
if (j == 0 || j == i - 1) {
smallRes.add(1);
} else {
// 这里错写过j-1
List<Integer> last = res.get(i - 2);
Integer count = 0;
int a = last.get(j - 1);
int b = last.get(j);
count = a + b;
smallRes.add(count);
}
}

res.add(smallRes);
}
return res;

}

static public List<Integer> getRow(int rowIndex) {

List<Integer> list = new ArrayList<Integer>(rowIndex);
if (rowIndex == 0) {
list.add(1);
return list;
} else if (rowIndex == 1) {
list.add(1);
list.add(1);
return list;

}

for (int i = 1; i <= rowIndex; i++) {
list.add(1);
for (int j = i - 1; j > 0; j--) {
// 只用到上一行的内容，set与add
System.out.println("j=="+j+"=="+list.get(j)+"=="+list.get(j-1));
list.set(j, list.get(j) + list.get(j - 1));
}
}
list.add(1);
return list;
}

static public String reverseWords(String s) {
// 时间复杂度结果为O(n^2)
// ""输入不能通过
if (s.length() <= 0) {
s = new String();
return s;
}
// 全部是空格的不能通过，去掉多余空格
int end = s.length() - 1;
int begin = 0;
while (begin <= end && s.charAt(begin) == ' ') {
begin++;
}
while (begin <= end && s.charAt(end) == ' ') {
end--;
}
if (end < begin) {
System.out.println("---s-");
s = new String();

return s;
}
// StringBuffer反之Time Limit 发生
StringBuffer res = new StringBuffer();
// 中间存储
String mid = new String();
int n = end;

while (n >= begin) {

if (s.charAt(n) == ' ') {
int smallC = mid.length() - 1;
System.out.println("===="+mid.length());
while (smallC >= 0) {
res = res.append(mid.charAt(smallC));
smallC--;

}
//依靠mid.length()去除中间多出的空格
if (mid.length() >= 1) {
res.append(' ');
}

mid = new String();
} else if (n == begin) {
//依靠空格判断做一次mid顺序置换，如果到最后一个单词的时候没办法处理，专门设定一个n==begin判断
System.out.println("====="+s.charAt(n));
mid = mid + s.charAt(n);
int smallC = mid.length() - 1;
while (smallC >= 0) {
res = res.append(mid.charAt(smallC));
smallC--;
}
} else {
mid = mid + s.charAt(n);
}
n--;
}
return res.toString();
}

static public boolean isHappy(int n) {

ArrayList array = new ArrayList();
// System.out.println(Integer.MAX_VALUE);
while (true) {
if (array.contains(n)) {
return false;
} else {
array.add(n);
}
n = count(n);
if (n == 1)
return true;
}
}

// 计算每一位的平方和
static public int count(int num) {
int count = 0;
int a;
while (true) {
a = num % 10;
num = num / 10;
count += a * a;
if (num <= 0) {
break;
}
}
return count;
}