leecode每日一练

打家劫舍

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我一开始的思路也是dp，但是转移方程想错了，这个题目转移方程应该是dp[i] = max(dp[i-2]+nums[i],dp[i-1])

class Solution {
public:int rob(vector<int>& nums) {int len = nums.size();vector<int> dp(len);int ans = 0;if(len>=1)dp[0] = nums[0],ans=max(ans,dp[0]);if(len>=2)dp[1] = max(nums[0],nums[1]),ans=max(ans,dp[1]);for(int i=2;i<len;i++){dp[i] = max(dp[i-2]+nums[i],dp[i-1]);ans = max(ans,dp[i]);}return ans;}
};

拆炸弹

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纯模拟就是sb，我用了差分数组，但是转移方程写死我了
我写的代码

class Solution {
public:vector<int> decrypt(vector<int>& code, int k) {int len = code.size();vector<int> ans(len);vector<int> p(len+1);p[0] = code[0];for(int i=1;i<len;i++){p[i] = p[i-1]+code[i];}if(k==0){return ans;}else if(k>0){for(int i=0;i<len;i++){if(i+k<len){ans[i] += p[i+k]-p[i];}else{ans[i] += p[len-1]-p[i];int cha = k-(len-1-i);if(cha>0)ans[i] += p[cha-1];}}}else{k = -k;for(int i=0;i<len;i++){if(i>k){if(i-1-k>=0)ans[i] += p[i-1]-p[i-1-k];else{ans[i] += p[i-1];}}else{if(i>0)ans[i] += p[i-1];int cha = k-i;ans[i] += p[len-1]-p[len-1-cha];}}}return ans;}
};

但是其实只要开两倍的空间就可以简单完成,下面这个思路是滑动窗口，但是其实差分也是可以的

class Solution {
public:vector<int> decrypt(vector<int>& code, int k) {int n = code.size();vector<int> res(n);if (k == 0) {return res;}code.resize(n * 2);copy(code.begin(), code.begin() + n, code.begin() + n);int l = k > 0 ? 1 : n + k;int r = k > 0 ? k : n - 1;int w = 0;for (int i = l; i <= r; i++) {w += code[i];}for (int i = 0; i < n; i++) {res[i] = w;w -= code[l];w += code[r + 1];l++;r++;}return res;}
};

基站建设

题目地址
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一开始我的思路是差分，但是只能过五分之一的数据集，这是因为如果这些重叠的区域过多的话就会出现问题

给你看一下差分的代码

#include<bits/stdc++.h>
using namespace std;const int N = 1000005;
int s[N];
int p[N];
int n;int main() {cin >> n;int a, b;int mm = 0;for (int i = 1; i <= n; i++) {cin >> a >> b;int l = (a - b) > 1 ? a - b : 1;int r = (a + b) < N ? a + b : N - 1; mm = max(mm, r);s[l] += 1;s[r + 1] -= 1;}for (int i = 1; i <= mm; i++) {p[i] += p[i - 1] + s[i];}int re = 0;int now;int flag = 0;for (int i = 1; i <= mm; i++) {//cout << p[i] << " ";if (flag == 0 && p[i] > 1) {flag = 1; // 进入区域now = p[i];}if (flag == 1 && p[i] <= 1) {flag = 0; //出去re += now - 1;}if (flag == 1 && p[i] > 1) {now = max(now, p[i]);}}cout << n - re;return 0;
}

应该用贪心来做，就是一个活动安排问题

#include<bits/stdc++.h>
using namespace std;const int N = 1000005;
struct node
{int start, end;bool operator<(node b) {return this->end < b.end;}
}a[N];int n;
int main() {cin >> n;for (int i = 1; i <= n; i++) {int x, b;cin >> x >> b;a[i].start = x - b;a[i].end = x + b;}sort(a + 1, a + 1 + n);if (n == 0) {cout << 0;return 0;}int ans = 1;int now = a[1].end;for (int i = 2; i <= n; i++) {if (a[i].start <= now) continue;now = a[i].end; ans++;}cout << ans;return 0;
}