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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 F8 v _8 I/ s4 I }
- if (sentence == null || sentence.length() == 0) { . T3 S+ `9 w; j7 p c
- return false; - }9 o0 N8 ?' e5 f7 h* p4 `
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- Set<Character> set = new HashSet<>();
9 l3 I# m! Z; o0 P& K - for (char c : sentence.toCharArray()) {
' P! Z, o0 C- M, Q. E - if (c - 'a' < 0 || c - 'a' >= 26) { " b" S, @# V+ ?
- return false; 9 F8 m6 L& v G1 ^0 p1 @- b
- }
0 z5 Y6 [4 U$ |% p8 x* u7 w - set.add(c);
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4 ?' e. I/ a* O - return set.size() == 26;
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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) { 7 m o" {! F) R- s# p
- if (costs == null || costs.length == 0) { 4 z8 \* G* R. S/ B% F4 S$ }
- return 0; $ Q9 v$ a# Z- d: H& b1 r
- }
) b& K2 q0 `3 N, X' B; q$ a0 ~6 A - int n = costs.length;
) U2 d2 \. a7 X3 r - int res = 0; % [4 \$ S2 ?# |" K) m
- // 排序从小到大买即可
, |& N$ j: D4 {: M0 Y+ ] - Arrays.sort(costs); # K" }8 W0 p% l) ?1 `3 g- ?' ~
- for (int cost : costs) { - o$ Y* a% d' B$ A
- if (coins > cost) { 8 d9 v+ x4 ~8 v5 @2 [% e
- res += 1;
X( v$ l& S1 Q" L) f - coins -= cost;
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- else { - a5 ~; U6 y+ v
- return res; . K( M2 {) D+ L1 X. W5 X
- }
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D0 u5 Y# V1 s1 i - return res; 3 L! B! A, ^ H
- }
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