[size=1em]想上岸 找上岸 No.1 Check if the Sentence Is Pangram 解题思路 简单遍历字符串判断 非英文字母字符直接返回false 否则记录不同字符的个数是否满足26个 代码展示 - public boolean checkIfPangram(String sentence) { , I, Z/ J* E+ r- ^- Q8 J$ ?
- if (sentence == null || sentence.length() == 0) { 8 o' `7 k% @4 K
- return false;
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- Set<Character> set = new HashSet<>();
8 Y5 k9 Z& h/ G7 { - for (char c : sentence.toCharArray()) {
0 j. s, w& T. N/ M - if (c - 'a' < 0 || c - 'a' >= 26) {
2 ]: P- I) X6 R0 ^ d7 h - return false; : h5 U% {7 G' b# G
- }
5 t$ q0 v! T2 q0 k- l3 Z4 k - set.add(c);
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- return set.size() == 26; 1 v8 c6 s! B. f9 G2 a4 J% L
- }
复制代码No.2 Maximum Ice Cream Bars 解题思路 贪心的去买雪糕 无需看成背包问题,TLE or MLE 代码展示 - public int maxIceCream(int[] costs, int coins) { & T6 @# e' O, ^ u$ C' }4 p
- if (costs == null || costs.length == 0) {
* q* U7 z: n0 L" _ - return 0; , [) ?( v- q- x% W8 [% J; k
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- int n = costs.length;
5 K" A. o- ~+ ^+ r. F; [% w4 P$ ~) a - int res = 0; ! u/ P+ ?; T* n( d2 d8 G2 x" M5 J+ y1 L
- // 排序从小到大买即可
) G) C6 q6 V( m2 V# p A0 L. w - Arrays.sort(costs); " g, r1 ^% O! ~5 |4 i" W& s& C
- for (int cost : costs) { 4 ]5 a# d: L, U# i1 ~5 U7 _& H" p
- if (coins > cost) {
: K! Y/ W% A( A# P2 M' f4 m( j - res += 1;
* d8 b& L n1 m% M - coins -= cost; & e" t* s2 U2 U
- } 8 w: f$ @$ h ~' b' d0 z& X
- else {
* t) w2 R" B2 j* d$ F. j. h; M/ y - return res; : R, P. [ m# M$ T/ S9 U- I: w) N O
- }
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- return res; 0 [# f$ e0 \! W$ {) n
- }
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