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[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) { 1 ~3 E: C: i9 P$ [5 Q; s
- if (sentence == null || sentence.length() == 0) {
" ~% k1 c) t% b/ K - return false;
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4 b: u, o' N1 _% B - Set<Character> set = new HashSet<>(); # k. n/ R( o' j2 f9 Q- y
- for (char c : sentence.toCharArray()) { 7 v Z* k, M2 G8 y. A
- if (c - 'a' < 0 || c - 'a' >= 26) { ! e F: ~; \0 x
- return false;
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- set.add(c);
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- G0 [# ]& y; ]7 Z$ e - return set.size() == 26; % S4 T- |- _. q. v
- }
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) { # h, k* Y) s" h: w: g5 a3 v- K( W. c
- if (costs == null || costs.length == 0) { - ~, o/ w5 e& [- u3 w
- return 0;
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8 r9 P; A- z y7 z1 [) a4 p' W - int n = costs.length; # |+ Q, o$ m% k$ V0 G8 f
- int res = 0;
4 g) Y% I# x% q8 d+ f! L - // 排序从小到大买即可 7 l9 ?" T, j0 |8 }8 s7 A' L4 M
- Arrays.sort(costs); 7 _7 c: z2 z2 q3 ?2 T2 u# s7 O
- for (int cost : costs) {
. s4 w2 d/ ~0 d$ a9 ` - if (coins > cost) { " S% K6 |, T5 i9 c3 c) `8 H3 W
- res += 1;
5 ~) ~) C2 U2 } P/ O - coins -= cost;
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. E: [, n* _7 ^& D. O& ^ - else { g0 S4 W" p6 y) Z0 k
- return res; 3 O' ]: F; t7 v' O2 \- m
- }
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- return res; ! ?! s5 u3 }% v; d
- }
复制代码/ y* n/ [0 d! y" V3 f; z8 ~
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