[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) {
0 `7 Z* R* s; w8 v8 _ - if (sentence == null || sentence.length() == 0) { 8 U# o" @" S8 F' l6 ^* v+ a
- return false;
* K2 p$ ?0 y) \/ A/ g - }
: R" _+ ]% ]. f3 Q& x; ` - Set<Character> set = new HashSet<>();
2 Y+ I$ W+ |& M. V' O - for (char c : sentence.toCharArray()) {
, L1 A3 ~! P/ P+ A0 @ - if (c - 'a' < 0 || c - 'a' >= 26) {
; `* K# D' \7 w0 M0 k: i - return false;
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- set.add(c); ) ^2 X& X! k& u% u7 h' }1 U
- }
* N. K2 P! G, T- k - return set.size() == 26; 1 s0 F* i' G' w9 ?6 Z; j
- }
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) { 3 y: I% T; l7 T8 o+ X5 N) E
- if (costs == null || costs.length == 0) {
; c& e/ D, E- }0 t6 T# E - return 0;
6 A8 {7 J# b" x% p8 C7 n - }
$ C! n' U3 c0 c0 Q$ R - int n = costs.length; ]" q0 N( p3 d) @
- int res = 0;
0 C' V+ n* t& E+ C4 C- B- [4 b( g/ g - // 排序从小到大买即可
) A) ?$ M! w5 C* g2 i# E- D - Arrays.sort(costs);
5 I; R. s% R1 |$ f. q2 q. x - for (int cost : costs) {
7 {) y' Q# |( @! y - if (coins > cost) { 0 J, ~) v5 \/ ~2 L+ ]1 m
- res += 1; 9 f( `& k- p s
- coins -= cost; * m; j* I# e8 W- Y. o( `
- }
- W9 d2 H; }1 s, l8 d* H) K% e - else { 1 j) O! @6 w, _2 G, e- q" v
- return res; # q& {* j2 q6 N7 o5 H c$ Y$ F
- } # _9 O( s: ]+ {$ |* i
- }
8 U: S5 { R) S. ^+ |1 j - return res;
9 E" @3 p6 C, z) l - }
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