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【 NO.1 统计数组中峰和谷的数量】. z5 [3 P P+ u0 n$ s1 @5 e
& h$ S, p% m0 U( v2 }4 }解题思路" ], s6 W }! p% Q
先去重,再统计。
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代码展示: m! J- [2 u# R7 u N8 Z
" H/ Q* e' W$ Q O8 o$ eclass Solution {
& L% k. V2 a. R public int countHillValley(int[] nums) {/ W1 v7 ^% g9 l& I5 F) r
int len = 1;3 S. k9 a3 u$ @7 b
for (int i = 1; i < nums.length; i++) {
. G6 y9 g. C! S! T9 S if (nums[i] != nums[len - 1]) {( U# |$ d; \7 D9 o' H, f/ q6 Q
nums[len] = nums[i];- }, o( a z1 T, G; }/ j- P$ L. y
len++;
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int res = 0;/ G* O- P( o l1 V" I: J( d* ?
for (int i = 1; i < len - 1; i++) {
1 I; f) \2 P3 k if (nums[i - 1] < nums[i] == nums[i + 1] < nums[i]) {
' [' J+ c: x8 V- j& C7 ]3 l7 o res++;1 W/ Q3 h0 K; f4 q) P2 v9 n
}
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return res;
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}
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【 NO.2 统计道路上的碰撞次数】
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5 y' J( B5 p( A解题思路
: e" i$ F% w- G) S从左到右遍历,维护左侧的车的状态,有两种:- M( d( b/ u$ q, g0 r, P
! L8 _7 _* ]) L: ?. {4 w
若干个向右行驶的6 t* a: x) b7 G1 B5 X1 y& ]
+ h, G3 T( K$ K# _5 l9 b2 K/ o停止
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# ]( d$ J! G8 _2 `代码展示
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5 B0 n m- X- c" ?/ Rclass Solution {9 ~) p7 g. A5 w
public int countCollisions(String directions) {
" v9 @; i- k% e }+ N1 U; ] int res = 0;
# S+ j* B- V& q int S = 0, R = 0;8 C7 O$ J& e6 c
for (char c : directions.toCharArray()) {1 O2 H, A$ n: W1 G
if (c == 'L') {
4 V& J% E6 L* t7 i if (S + R == 0) {! p% N8 M1 K: J0 `* J
continue;- P0 k& P1 ~# H% d
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if (R > 0) {
S, }) Q# `6 i. ]6 h" X7 Y res += R + 1;
2 d7 k/ P7 {& {# }, [ } else {
8 {1 l: U# X6 @9 @ res += S;
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S = 1;( u# E# ?8 P- I( v n1 F
R = 0;
3 C1 G3 D5 B" N6 J }
9 ?! e z; @& K9 z6 h0 E% e: D if (c == 'R') {
5 `+ g- O- B9 u4 l S = 0;# R2 W1 P1 }+ I; F" K
R++;- `1 D! T4 u& K7 s- {# X3 E
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if (c == 'S') {
) ~ _; H, M9 I) T0 J res += R;
$ e/ D! M6 v/ T0 e S = 1;
2 C9 t+ c4 ?) c% d# G$ |6 d4 z- R R = 0;: y& R* _0 J( C7 ~- u/ B3 m! \, p' ^
}
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5 {8 {+ a n2 C: `- E return res;- v# Z/ o5 L/ X! g, r% I
}
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【 NO.3 射箭比赛中的最大得分】; g m2 L1 @0 y; x0 c6 Z, J' A
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解题思路- Z, L; ~6 j6 D$ G! s5 v; F7 L
枚举所有的情况即可,实际一共有 11 个得分位置,枚举 Bob 要在哪些位置得分。) e$ f7 E, J: S! ]
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贪心地,当 Bob 要在位置 i 得分时,只需要花费 aliceArrows[i] + 1 支箭。
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最后多余的箭可以放到位置 0.
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4 X; N7 k9 \) x# {代码展示
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- D2 q2 L& J# V( E" Xclass Solution {8 V; d2 a& u' ]! s8 s8 j
public int[] maximumBobPoints(int numArrows, int[] aliceArrows) {
. Y& ?) P$ P; k# x! X int max = 0;
/ a5 _. H$ {1 H) _, `) U int[] res = new int[12];. s6 x3 H T6 j& X. @+ X
for (int i = 0; i < (1 << 11); i++) {
3 a- a' ]$ f* v int num = 0;" ]7 y6 |; Z) ~
int score = 0;0 g& C! \" [, q! z. J' R
for (int j = 0; j < 11; j++) {' z6 Y0 Y! k+ T* q/ r. y; B& i) A
if ((i & (1 << j)) > 0) {
' \$ G. [- C% e9 @- I8 J* y num += aliceArrows[j + 1] + 1;
1 b- o2 u4 ^. l3 ?% ?6 D) T: H score += j + 1;
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}
" C7 E: J5 p. P; M/ J if (num > numArrows) {
4 G9 t. H6 d( O: w$ n; i continue;/ n7 I; \' b! ^/ @6 i4 O: {9 p! p
}
, y& p7 l8 N7 M0 h! `) K9 X& p if (max < score) {
4 @* s7 G% f" v1 ^ _* @3 J7 H2 Y& n max = score;
+ g% h- H, y4 X6 l4 P+ b res[0] = numArrows - num; // 多余的箭& u& h$ g0 s' z* s8 ^" u
for (int j = 0; j < 11; j++) {
; s& O8 o) j0 n! u n if ((i & (1 << j)) > 0) {0 }! D# j4 y& N" m
res[j + 1] = aliceArrows[j + 1] + 1;
* R: I* ?+ {3 ]. a- F3 I% ^ } else {
2 k" J( e2 r" w" z7 t4 |( M res[j + 1] = 0;
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}
1 a/ P' w- U; b7 y; h3 g return res;
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}
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【 NO.4 由单个字符重复的最长子字符串】- `$ b1 Y: @: N5 K, x9 z: `0 y' J
6 b* B) Q6 z2 |) Y" D& W解题思路2 G1 S8 D2 T4 ?. u: ^0 s& m9 R
灵活运用 TreeSet 即可,详见注释。- n- T; C: t* A% T! y
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代码展示
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class Solution {8 u; v3 A. N$ ~% s: r
2 m3 }! F+ ?7 o) I/ P4 Y // 一个 Interval 表示一段连续的字符
/ r, G3 Z6 s: q% [* v& U4 B static class Interval implements Comparable<Interval> {7 |0 b r7 s8 O$ W1 d* h
int start;8 x( ]% q+ Q/ }7 a+ Y
int len;
% C) b, r) v: n8 |' ]" u' p+ D char c;
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public Interval(int start, int len, char c) {
- s! I/ q" i) H i) w! P4 O this.start = start;+ S/ e! b6 N. t; s/ R& |+ D
this.len = len;
! {7 \3 c7 r; T7 V( C this.c = c;6 {6 P9 x4 N+ Y
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public int compareTo(Interval o) {
& j; K" t$ V6 j3 T4 P3 e if (len != o.len) {! M/ ^; h. h* k7 g% d
return len - o.len;+ @- i p" ^: _1 E# g
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if (start != o.start) {
8 a5 Q3 [: m3 Q' c" z: w# P return start - o.start;
* F7 d8 R% Y$ r }
3 L# B0 J# c2 i. X# {' t return c - o.c;/ T7 O4 K( @" s3 K4 |
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}
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public int[] longestRepeating(String s, String queryCharacters, int[] queryIndices) {/ A6 Y# G( n0 ~. P3 l) c
// 首尾新增两个其他字符,这样相当于 queryIndices 不会修改字符串首尾,便于后续的处理
! X; ^5 q# ` G* Y s = "_" + s + "_";
( l5 h; h: J% Z" |5 `! a' U // allIntervals 按照 len 排序全部的区间" M% i. J& {. B, ~4 @8 K9 \
TreeSet<Interval> allIntervals = new TreeSet<>();5 ?; l5 S: Q& g H% B) J
// intervals 按照字符聚合区间,即 intervals.get('a') 表示字符 a 的全部区间,并且按照起始位置排序! w% r- u" p' }% A( r6 \: J) W
Map<Character, TreeSet<Interval>> intervals = new HashMap<>();+ N/ R6 n& K* o% r1 r! a& C
for (char c = 'a'; c <= 'z'; c++) {
2 m5 z3 W: q$ ^4 b intervals.put(c, new TreeSet<>(Comparator.comparingInt(o -> o.start)));
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intervals.put('_', new TreeSet<>());3 e8 ^; Y4 T2 u2 a2 D, e1 @* `
6 I s4 [& [' f: G, d // 遍历 s 维护 intervals 和 allIntervals
4 {6 t' F T& _ int last = 0;
% h0 n. m7 O1 k( l! t0 W' t8 | for (int i = 1; i < s.length(); i++) {, T" v; m& U. b$ z, K4 k' K
if (s.charAt(i) == s.charAt(last)) {/ \1 V$ G! M* @
continue;
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Interval interval = new Interval(last, i - last, s.charAt(last));2 H; k# K* g" c' `1 w
allIntervals.add(interval);
1 B* g2 @/ B m! ~! e intervals.get(s.charAt(last)).add(interval);
/ b; D5 x$ j8 a% O- [- d; _ last = i;
& z, f1 z% X6 S2 X4 p }
, C6 R9 Q8 _2 x" P Interval interval = new Interval(last, s.length() - last, s.charAt(last));
4 ~3 O8 H# ^$ v3 L. O; ^ allIntervals.add(interval);8 c6 t7 U$ ^/ r5 |. p
intervals.get(s.charAt(last)).add(interval);6 P8 W! ~# p7 W, M k
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// 每次 query 调用 maintain 维护 intervals 和 allIntervals$ i. g5 k- o% D6 U7 }
int[] res = new int[queryIndices.length];7 o- ?# i$ T2 g$ r7 E; L
char[] arr = s.toCharArray();
3 v1 [! C) @+ }/ s; M3 `6 W for (int i = 0; i < queryIndices.length; i++) {
1 y2 \8 I+ m4 c& h4 A7 _. g! b res[i] = maintain(arr, allIntervals, intervals, queryIndices[i] + 1, queryCharacters.charAt(i));
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return res;9 u, [- f. u8 R, i
}
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// 将 arr[idx] 替换为 c
- w' f: B! a1 O2 }+ ^( J* ~ // 维护 allIntervals 和 intervals
/ u7 S$ P- T$ M# r/ x) q$ R // 返回单个字符的最长的子字符串长度1 K2 _$ q' X: g& Z3 m5 A8 B y' C
private int maintain(char[] arr, TreeSet<Interval> allIntervals, Map<Character, TreeSet<Interval>> intervals, int idx, char c) {
/ z2 [+ Y! U$ c: G if (arr[idx] == c) {
) B8 s J8 I- j5 a return allIntervals.last().len;) D9 @! @+ v0 g* t, h
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* T9 ]0 v9 M$ U! P7 L/ r // 维护原字符 arr[idx] 的 interval
$ n1 v1 B9 ?0 t$ d% o H var treeSet = intervals.get(arr[idx]);7 F3 H+ o1 k1 f* B b
Interval origin = treeSet.floor(new Interval(idx, 0, arr[idx])); // 调用 treeSet.floor/ceiling 时,只需要 start 即可$ r! j; d8 @5 V* X8 F+ H/ _. e* _
treeSet.remove(origin);
) A% D j! u( d6 D9 Z) y; P allIntervals.remove(origin);
( i5 w, Y% e D5 u. i if (origin.start < idx) {4 c7 `6 M2 G( B4 o' D9 D" p2 o. O' R/ d
Interval interval = new Interval(origin.start, idx - origin.start, arr[idx]);
4 u2 K! q& H' [5 |' P2 A treeSet.add(interval);: o7 K0 \2 \- b5 s, e' K
allIntervals.add(interval);
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; ^2 S q! [" V* J. o if (idx + 1 < origin.start + origin.len) {
/ c: c" z- N: w9 U( N Interval interval = new Interval(idx + 1, origin.start + origin.len - idx - 1, arr[idx]);, E6 F1 E o* Z( S
treeSet.add(interval);
" X2 |$ U6 W# i. m allIntervals.add(interval);
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- h8 y- @/ `5 |+ N2 D5 R9 N // 维护新字符 c 的 interval
7 V: v0 J3 s( e0 N( m5 y: a treeSet = intervals.get(c);
- x; W$ ]; ~- X: e# K/ d5 C5 S if (arr[idx - 1] == c && arr[idx + 1] == c) { // 左右连接
+ t4 X5 Q8 f ~, h6 D Interval left = treeSet.floor(new Interval(idx - 1, 0, c));
& q3 d2 n# ^0 ^0 n Interval right = treeSet.ceiling(new Interval(idx + 1, 0, c));7 k( R: V \. Q4 ^
treeSet.remove(left);) e2 o$ K* l# ]- g8 F2 o: I
treeSet.remove(right);
8 ~9 `( x1 Y/ d. A0 e6 J allIntervals.remove(left);
6 ~; F6 z. G4 o; X allIntervals.remove(right);
[' @, i) \* K4 I( _ Interval interval = new Interval(left.start, left.len + right.len + 1, c);7 I! P3 _- l0 i. p: U
treeSet.add(interval);: x6 Q. N5 }# R
allIntervals.add(interval);
: w6 l$ ^& h/ k, Z } else if (arr[idx - 1] == c && arr[idx + 1] != c) { // 左连接6 e: k; e! S0 B* U0 R
Interval left = treeSet.floor(new Interval(idx - 1, 0, c));: s6 ]' h6 T9 X4 c2 L# I0 k" n! q
treeSet.remove(left);
6 |2 l9 l* Q9 o" v+ t: U) v: V0 ~* [ allIntervals.remove(left);8 r4 u+ E, K" J+ P G3 b
Interval interval = new Interval(left.start, left.len + 1, c);8 h% Q6 `* S& ^* V6 `' @7 _
treeSet.add(interval);
7 w ?- n) i( U allIntervals.add(interval);6 p1 [1 A* ]0 L- U" U
} else if (arr[idx - 1] != c && arr[idx + 1] == c) { // 右连接
( z; W2 `" n! M- N' n/ h# j: |5 R1 f Interval right = treeSet.ceiling(new Interval(idx + 1, 0, c));& T9 _8 E7 T0 x# s
treeSet.remove(right);: F) L9 I W, A6 Z( s) q3 [
allIntervals.remove(right);4 i# {: q; }* v8 m, v9 i/ r: g
Interval interval = new Interval(idx, right.len + 1, c);
( g# N/ T( W8 V' a treeSet.add(interval);
, k- D8 {5 I& J allIntervals.add(interval);; V8 S$ E6 w& k% B' ?4 j5 F ^% R
} else { // 单独一个- h; T1 N" }1 }& T" H! J
Interval interval = new Interval(idx, 1, c);
$ H; H" [, z- H' C$ M' J7 q7 w treeSet.add(interval);
. v0 Y2 }" Q/ F4 M" g: s9 t allIntervals.add(interval);% d; x, m8 M$ N7 X
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arr[idx] = c;
; E' z' o/ w5 q+ V- M a/ G return allIntervals.last().len;
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} |