|
[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) {
& x' r# y% O2 z+ y - if (sentence == null || sentence.length() == 0) { , u0 X& ]2 s0 w' g3 o# Q2 ?4 L
- return false; # N) Z+ D) K& j6 y/ f- K3 c8 |3 O
- }
9 ] w: P1 R, ?" I5 _& W# E0 Q' B - Set<Character> set = new HashSet<>();
+ E* A8 L: c- d1 \ - for (char c : sentence.toCharArray()) { ( p0 @$ R% a. P( u. U2 ^: k
- if (c - 'a' < 0 || c - 'a' >= 26) {
; K' A7 L) e+ ^, x, f) n7 M1 Y$ _ - return false;
/ A# p b6 y' H# a: y - }
* n* m. ^$ f- G8 F4 m - set.add(c); " o, v/ O4 Y' }6 `% C/ l& _+ A
- } ) X3 q. v! s/ x( c/ K) Y6 N! X
- return set.size() == 26;
% X7 u) c4 q& Y7 E8 \/ R- C: [1 ^ - }
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) { * |# ]5 h+ }0 k; }: W1 B( S) c
- if (costs == null || costs.length == 0) {
7 \) p' v1 \# g% d - return 0; 9 C$ z/ e4 W( T9 d5 e( V0 x& x
- }
; Z+ X9 }5 F& `9 z8 E3 ?7 J# i% r - int n = costs.length; 8 E- M$ [* p7 Y, W& E
- int res = 0;
! Q0 y" V" K' W- | T% ~ - // 排序从小到大买即可
! T9 p; |7 r) ?! i; y- C8 E& d1 m - Arrays.sort(costs);
) }) s- g7 @9 ~: G - for (int cost : costs) {
! X$ `# h- W$ x" S - if (coins > cost) { . n7 G; e/ D4 m0 U; h0 \
- res += 1; : Y* b8 e" ~( l! W
- coins -= cost;
8 ^" Z8 |2 U0 F. c$ |. ]2 ? - } ( T1 r" j; X2 a$ O4 |
- else {
: L5 _8 u2 a% Z7 W7 q" k: L - return res;
% B7 U8 e- W0 \0 Y" N. p - } 0 K0 {6 e; ?1 S+ Y) V- U
- }
' s0 _6 y( f$ P4 M! F2 Q9 N; x - return res;
! z; w8 g4 _% n# R - }
复制代码% |1 l+ O7 N5 p+ S+ S
9 e% l2 h7 ~# f3 `; R
. C3 |, Q0 q }( x; I: j; z
% \% q- |2 a+ Y2 ]0 i b: p. `6 u; J6 q! e3 `
|