【每日·两道·算法题】 —— 第一天

第一题

最大连续1的个数

$Java代码：$

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int count = 0, maxcount = -1;
        for(int i = 0; i < nums.length; i++){
            if(nums[i] == 1){
                count++;
            }else{
                count = 0;
            }
            //if(count > max) max = count;
            maxcount = Math.max(count, maxcount);
        }
        return maxcount;
    }
}

• Java中一个比较大小的函数：Math.max(num1,num2);在C++中为max(num1,num2)函数.
• 计算数组长度的函数：
  nums.length (Java)
  nums.size() (C++)

$C++代码：$

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {

        int maxcount = 0, count = 0;
        for(int i = 0; i < nums.size(); i++){
            if(nums[i] == 1)    count++;
            else    count = 0;    
            maxcount = max(count, maxcount);     
        }
        
        return maxcount;
    }
};

第二题

数组之提莫攻击

$C++代码：$

int findPoisonedDuration(vector<int>& timeSeries, int duration) {
    int counttime = duration;
    int length = timeSeries.size();
    for(int i = 0; i < length - 1; i++){
      
        // if(timeSeries[i + 1] - timeSeries[i] <= duration) counttime += timeSeries[i + 1] - timeSeries[i];
        // else counttime += duration;
        counttime += min(timeSeries[i+1]-timeSeries[i], duration);
    }
    return counttime;
}

• 获取最小值的函数

• 有时候判断大小比较的两种情况，可以用一条语句代替

if(a > b) sum += b;
else sum += a;
//可代替为:
sum += min(a, b);

$Java代码：$

class Solution {
    public int findPoisonedDuration(int[] timeSeries, int duration) {
        int time = duration;
        for(int i = 0; i < timeSeries.length - 1; i++){
            // if(timeSeries[i+1]- timeSeries[i] < duration) time = time + timeSeries[i + 1] - timeSeries[i];
            // else time = time + duration;
            time += Math.min(timeSeries[i+1]-timeSeries[i], duration);
        }
        return time;
    }
}