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【 NO.1 检查是否所有 A 都在 B 之前】8 w. V4 K* U& K$ v5 m$ D
* @+ H, M. c" h3 f, S解题思路. l& G, i( g& n% d
签到题。
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代码展示
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# \' ~- [, z0 E+ Gclass Solution {
- ]# ^( M, _# D public boolean checkString(String s) {% ` x1 s5 Z% B$ @: [% i# J; Q* R
return !s.contains("a") || !s.contains("b") || s.lastIndexOf('a') < s.indexOf('b');5 c/ S w6 U1 _ i
}
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【 NO.2 银行中的激光束数量】
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解题思路' {& k+ Y6 m" ~0 R' Y. v
统计每行 1 的数量即可,如果没有 1 则跳过。) O; I {' p# L" l
# S4 \! z' L: u/ K代码展示
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class Solution {
. _ ^8 l3 Y' H9 T z9 @: g public int numberOfBeams(String[] bank) {8 y) q% k6 b/ E3 N: R
int result = 0;, ]5 n, l7 W$ j5 h3 ]( A+ f W0 |
int last = 0;0 d o7 A; f1 w
for (var s : bank) {! ?- g- d# s' r( b @3 [6 \* w
int count = count1(s);( o0 f$ t; [: T7 [
if (count > 0) {' i( Z! D4 f8 d6 i% w
result += count * last;
$ r" H2 |0 L8 c1 g. M last = count;
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return result;
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4 R* u* d& N7 I! m* k) p5 ] private int count1(String s) {
& j9 k* v; [- k5 k int r = 0;. w p. k& V4 {* B1 }2 J! f
for (var c : s.toCharArray()) {. ^3 x& B; I3 H4 I
if (c == '1') {. G$ m) i! u2 J1 S+ y4 ~3 w$ e3 V
r++;; {/ P5 u; }! c* W9 U
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}
: v9 J; x$ r) l! o' Z7 o return r;( M9 m6 i+ i5 g
}
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【 NO.3 摧毁小行星】
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& }4 ` Z0 C+ r' Y$ h解题思路
+ M% h% p$ g$ Z0 Z排序即可,注意使用 long。
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代码展示; P: k2 n5 ~* S& i. H
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class Solution {
" K6 A. C" b/ G% h# T public boolean asteroidsDestroyed(int mass, int[] asteroids) {
, _+ l8 h7 n* e/ { Arrays.sort(asteroids);
, x" h- @" ^- ]; N long m = mass;
4 r( }1 |0 b, T& i( G- C2 c8 p6 A for (int a : asteroids) {
V- W$ L4 X2 Y: E6 g( Q if (m >= a) {3 g6 ], J! ~3 F5 E8 b9 v
m += a;; t) O! f( t' G$ J
continue;4 ~/ S3 |2 n- k( a; x+ q
}
4 q$ I9 ]3 {" i, I$ v F2 a/ }2 l return false;
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return true;- S& ^" ` ^7 y4 s
}
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$ }; w, L) g3 W1 \【 NO.4 参加会议的最多员工数】
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5 p. n9 A6 R' Q+ H解题思路
, i7 K+ {" E' T% ]7 y& C+ n实际参加会议有两种情况:; Q& a& z1 x4 n) E
3 T# w- \1 g. R1 S: K刚好是一个环中的人参加会议,这样可以使得每个人都喜欢自己左侧的人。1 Y& ]1 q6 Q2 l
+ B; Q0 s& @6 ~) r% b6 t% y有两个人相互喜欢,然后剩下的人形成一条链,比如 [1,0,0,2,1,4,7,8,9,6,7,10,8],这样不要求首尾相接,可以有多条链。) s. b4 z# U/ }5 x: T! R8 G1 f0 G
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2 \3 D" g6 M4 W+ |代码展示
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class Solution {. e6 |8 {' u6 U7 ^
public int maximumInvitations(int[] favorite) {
7 [' X$ N4 e0 V4 D4 d8 q$ Y, g/ O // 1 找一个完整的环
. }& k8 | p0 }5 j5 c+ H5 f0 e) z // 2 找多条链# z4 Z* z. Z" s& P! q- V' e, O
return Math.max(maxCircle(favorite), multipleLink(favorite));- \6 Q) B$ Z4 i, g* ~; X
}
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3 }* ?* {4 t% Z% S% a" L8 Z private int multipleLink(int[] favorite) {
2 N( x4 p$ k+ D$ Y3 C: r( L. } Map<Integer, List<Integer>> reverse = new HashMap<>();$ a U9 n; n, B; w5 A8 i
for (int i = 0; i < favorite.length; i++) {
6 G, ?# ]2 a" R" C" ] int from = favorite[i];% {! `% _+ y3 ^' X( n
int to = i;
; c2 b2 Y2 h. Y; j0 N if (favorite[from] == to) {
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if (!reverse.containsKey(from)) {. C. `; j' y& F- q/ P
reverse.put(from, new ArrayList<>());7 {" k1 Q) C" ]8 G2 Q; \
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reverse.get(from).add(to);0 C: c: F2 r% U" S/ y/ B
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int result = 0;6 q5 ^" @' T$ X |* `
for (int i = 0; i < favorite.length; i++) {
9 i' _9 l' L Q- ~0 C% s0 x if (i < favorite[i] && favorite[favorite[i]] == i) {# Z& ^4 P5 W8 A. f8 q. Z3 x
result += maxLen(i, reverse) + maxLen(favorite[i], reverse);, g7 X- u: W, e- ?6 s0 u& o+ \
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}
6 Q1 z! m1 M9 ]7 E% D4 H, Y return result;+ L0 @& b, T, f8 q$ y" w) o
}
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private int maxLen(int start, Map<Integer, List<Integer>> reverse) {+ {/ M# Q6 Y+ Q6 ^# x3 ?* P$ b
if (!reverse.containsKey(start)) {* j9 w$ I" ]9 X3 f, z- H
return 1;, J$ h! m3 |- M9 G1 H: L2 o4 L' d
}
: x8 l& _+ [* j# J& {( N+ ?+ h' [9 ] int max = 0;" R# c- ]! n5 s, s0 c3 B# D9 K p
for (int next : reverse.get(start)) {
) N' u" e/ e4 V1 e: q max = Math.max(max, maxLen(next, reverse));- C" z* O m% h
}
% @2 R& U+ c8 N" l# o6 U return max + 1;
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; P& s" w. w0 V' z private int maxCircle(int[] favorite) {. ^, I' y# c# q" `, R6 r
Map<Integer, Integer> step = new HashMap<>();
+ a% p2 }$ F6 N7 e6 y% e T* P int[] max = new int[favorite.length];
# M. J; G3 p( n int res = 0;4 l. ~8 c. [: t# O
for (int i = 0; i < favorite.length; i++) {9 f) B: s9 `4 r- I9 w* U
if (favorite[favorite[i]] == i) {- _. R% v8 S9 e1 Q/ p$ m9 u0 J4 @3 z' Y6 _
max[i] = max[favorite[i]] = 2;
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F( ?5 {. T4 t$ o" Z7 G. c if (max[i] == 0) {7 R* f. C6 d$ h- H
step.clear();
, Q q& ~, C) q3 }+ o2 F getMaxCircle(0, i, step, max, favorite);
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! X$ A( l! t. [9 I9 W9 ] res = Math.max(res, max[i]);
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return res;& R; ~2 z3 J! Y6 X; n
}
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- z4 \8 ?% S1 a" c private void getMaxCircle(int i, int cur, Map<Integer, Integer> step, int[] max, int[] favorite) {
3 Q8 M! j9 [6 O0 g if (step.containsKey(cur)) {$ [ i- [4 i k, e) [) y" g
max[cur] = i - step.get(cur);
9 A8 R0 X- _2 [ return;
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: i2 B' v& N( t* T% ]2 o step.put(cur, i);" [& O1 T2 \4 f0 T" E
int nxt = favorite[cur];. L2 \! y3 O% c [
if (max[nxt] > 0) {
{4 k7 ]9 m: U C max[cur] = max[nxt];
$ @; S! Z) x% G* @ return;
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getMaxCircle(i + 1, nxt, step, max, favorite);
5 S. [7 e' w5 Z3 w max[cur] = max[nxt];- m3 z/ H/ F3 f$ T' ~# E
} _8 w/ E: W4 u! r$ a3 ~
} |