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Commit 71a611a

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20190324
1 parent 2d9f477 commit 71a611a

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9 files changed

+171
-2
lines changed

9 files changed

+171
-2
lines changed

‎code/lc15.java

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* 难度:Medium
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* 分类:Array, Two Pointers
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* 注意:如何避免 List 重复元素
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* Tips:lc15, lc16, lc923
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*/
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import java.util.*;
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‎code/lc16.java

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* 难度:Medium
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* 分类:Array, Two Pointers
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* 思路:3sum的思路,每次记下最接近的res即可
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* Tips:
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* Tips:lc15, lc16, lc923
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*/
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public class lc16 {
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public int threeSumClosest(int[] nums, int target) {

‎code/lc53.java

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* 难度:Easy
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* 分类:Array, Divide and Conquer, Dynamic Programming
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* 注意:分治方法如何进行merge,merge时,必须包含mid元素,因为是连续子序列
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* lc978
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*/
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public class lc53 {
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public static void main(String[] args) {

‎code/lc560.java

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/*
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* 560. Subarray Sum Equals K
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* 题意:子数组的和等于k的个数
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* 题意:连续子数组的和等于k的个数
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* 难度:Medium
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* 分类:Array, Hash Table
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* 思路:求出累加和存在hashmap中,如果当前hashmap中存在sum-k,那么就是一个解

‎code/lc921.java

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package code;
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import java.util.Stack;
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/*
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* 921. Minimum Add to Make Parentheses Valid
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* 题意:最少添加符号个数,使得表达式有效
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* 难度:Medium
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* 分类:Stack, Greedy
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* 思路:匹配的都出栈,最后剩下的栈中没匹配的个数,就是需要添加的个数
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* Tips:
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*/
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public class lc921 {
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public int minAddToMakeValid(String S) {
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char[] ch_arr = S.toCharArray();
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Stack<Character> st = new Stack<>();
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for (int i = 0; i < ch_arr.length ; i++) {
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if(ch_arr[i]==')' && !st.isEmpty() && st.peek()=='('){
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st.pop();
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}else{
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st.push(ch_arr[i]);
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}
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}
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return st.size();
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}
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}

‎code/lc922.java

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package code;
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/*
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* 922. Sort Array By Parity II
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* 题意:奇数位置上都是奇数,偶数位置上都是偶数
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* 难度:Easy
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* 分类:Array, Sort
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* 思路:题看似简单,但最优的方法并不好写
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* Tips:
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*/
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public class lc922 {
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public int[] sortArrayByParityII(int[] A) { //O(N^2)
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for (int i = 0; i < A.length ; i++) {
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if( A[i]%2==i%2 ) continue;
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for (int j = i+1; j < A.length ; j++) {
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if(A[i]%2!=A[j]%2) {
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int temp = A[i];
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A[i] = A[j];
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A[j] = temp;
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break;
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}
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}
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}
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return A;
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}
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public int[] sortArrayByParityII2(int[] A) { //很巧妙的方法 O(N)时间, O(1)空间
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int cur = 1; //奇数位置的位置
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for (int i = 0; i < A.length ; i+=2) { //这里+2 只判断偶数位置上是否符合
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if(A[i]%2==1){
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while(A[cur]%2==1) cur+=2;
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int temp = A[cur];
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A[cur] = A[i];
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A[i] = temp;
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}
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}
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return A;
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}
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}

‎code/lc923.java

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package code;
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import java.util.Arrays;
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import java.util.HashMap;
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/*
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* 923. 3Sum With Multiplicity
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* 题意:3Sum有几种?数组中有重复数字
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* 难度:Medium
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* 分类:Two Pointers
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* 思路:由于本题只要求给出多少种Int值,所以不一定非要用3Sum的思路,有很多更简答的方法
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* 3种思路
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* Tips:
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*/
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public class lc923 {
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public int threeSumMulti(int[] A, int target) {
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int res = 0;
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HashMap<Integer, Integer> hm = new HashMap();
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for (int i = 0; i < A.length ; i++) {
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res = (res+hm.getOrDefault(target-A[i],0))%1000000007; // i之前的数字,两个组合起来,是否 == target-A[i]
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for (int j = 0; j < i ; j++) {
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hm.put(A[i]+A[j], hm.getOrDefault(A[i]+A[j], 0)+1);
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}
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}
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return res;
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}
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public int threeSumMulti2(int[] A, int target) { //3Sum的思路
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int res = 0;
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Arrays.sort(A);
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for (int i = 0; i < A.length-2 ; i++) {
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int left = i+1, right = A.length-1;
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while(left<right){
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if(A[i]+A[left]+A[right]<target) left++;
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else if(A[i]+A[left]+A[right]>target) right--;
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else if(A[left]==A[right]) { //如果相等,则直接 C N 取 2,计算出来,然后break
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res = (res + (right-left)*(right-left+1)/2)%1000000007;
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break; //不用继续移动指针了
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} else {
38+
int leftcount = 1, rightcount = 1;
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while(A[left]==A[left+1]) {
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left++;
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leftcount++;
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}
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while(A[right]==A[right-1]) {
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right--;
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rightcount++;
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}
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res = (res + leftcount*rightcount)%1000000007;
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left++; //别忘了,最后还要操作一下
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right--;
50+
}
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}
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}
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return res;
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}
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public int threeSumMulti3(int[] A, int target) { //直接数学计算
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long[] c = new long[101];
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for (int a : A) c[a]++;
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long res = 0;
60+
for (int i = 0; i <= 100; i++) // 题目给了值不超过100
61+
for (int j = i; j <= 100; j++) {
62+
int k = target - i - j;
63+
if (k > 100 || k < 0) continue;
64+
if (i == j && j == k)
65+
res += c[i] * (c[i] - 1) * (c[i] - 2) / 6;
66+
else if (i == j && j != k)
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res += c[i] * (c[i] - 1) / 2 * c[k];
68+
else if (j < k)
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res += c[i] * c[j] * c[k];
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}
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return (int)(res % (1e9 + 7));
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}
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}

‎code/lc978.java

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package code;
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/*
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* 978. Longest Turbulent Subarray
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* 题意:小,大,小,大 这种最大长度是多少
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* 难度:Medium
6+
* 分类:Array, Dynamic Porgramming, Sliding Windows
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* 思路:O(n) 遍历一遍就能解决,不用dp
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* Tips:lc53
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*/
10+
public class lc978 {
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public int maxTurbulenceSize(int[] A) {
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int inc = 1, dec = 1, result = 1;
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for (int i = 1; i < A.length; i++) {
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if (A[i] < A[i - 1]) { // +1 并且重置另一个统计量
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dec = inc + 1;
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inc = 1;
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} else if (A[i] > A[i - 1]) {
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inc = dec + 1;
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dec = 1;
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} else {
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inc = 1;
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dec = 1;
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}
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result = Math.max(result, Math.max(dec, inc));
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}
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return result;
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}
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}

‎readme.md

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| 104 [Java](./code/lc104.java)
9999
| 105 [Java](./code/lc105.java)
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| 108 [Java](./code/lc108.java)
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| 112 [Java](./code/lc112.java)
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| 113 [Java](./code/lc113.java)
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| 114 [Java](./code/lc114.java)
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| 116 [Java](./code/lc116.java)
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| 125 [Java](./code/lc125.java)
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| 127 [Java](./code/lc127.java)
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| 128 [Java](./code/lc128.java)
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| 129 [Java](./code/lc129.java)
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| 130 [Java](./code/lc130.java)
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| 131 [Java](./code/lc131.java)
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| 134 [Java](./code/lc134.java)
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| 148 [Java](./code/lc148.java)
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| 149 [Java](./code/lc149.java)
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| 150 [Java](./code/lc150.java)
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| 151 [Java](./code/lc151.java)
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| 152 [Java](./code/lc152.java)
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| 155 [Java](./code/lc155.java)
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| 160 [Java](./code/lc160.java)

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