Brush force buckle from 0

Write it at the front ：
Blogger Homepage ： To poke , Welcome the big guy to give directions ！
Blogger Qiu ：QQ：1477649017 Welcome like-minded friends to come on
Goal dream ： Enter the large factory , Determined to be a cow breaking Java Program the ape , Although he is still a little rookie now, hehe
----------------------------- Thank you for being so handsome and beautiful. Give me some praise ！ More than a heart -----------------------------

---

Recommend an article to plan the sequence of questions , If there is a little friend like me who doesn't know how to brush the questions , You can go to Kangkang ！ Poke here >>>

---

One , Array class

1.1, Traversal of array

Question no ：485, Maximum continuous 1 The number of

Subject portal , Poke poke began to solve the problem >>

---

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

This question examines the traversal of arrays and your ability to observe , Note that this is a binary one-dimensional array , As the tip says , Array element it is not 0 Namely 1, Then we should count the continuous 1 The number of , encounter 1 Just put count++, And synchronously save to max Inside , If the array element is 0 了 , Just put count Zero clearing , And continue to count again , Of course before max Or has been stored encounter 0 The previous continuous 1 The number of , Later, as long as it is updated in real time max Just fine .

---

Question no ：495, Timo attacks

Subject portal , Poke poke began to solve the problem >>

---

class Solution {
    
    public int findPoisonedDuration(int[] timeSeries, int duration) {
    
        int ret = 0;
        int end = 0;// Record the theoretical end time of poisoning 
        for(int i = 0;i < timeSeries.length;i++){
    
            if(timeSeries[i] >= end){
    
                ret += duration;
            }else{
    
                ret += timeSeries[i] + duration - end;
            }
            end = timeSeries[i] + duration;
        }
        return ret;
    }
}

---

In fact, this problem should not be too complicated , It is a problem of cumulative calculation of poisoning time , The only thing that needs our attention is to reset the poisoning effect . Every time we receive an attack , The time of poisoning will be duration, So let's not consider the overlapping of our time , First of all duration Add up , Then subtract the overlap from each time , That will finally be our actual poisoning time .
Put this idea on every attack ,end It records the end time of our theoretical poisoning ( Do not consider being attacked in the middle , as well as end This moment is not poisoned ), So our
timeSeries[i] >= end, Then there is no overlap , How many direct attacks , hold duration Just add up , Of course, every time end It will update , When not satisfied timeSeries[i] >= end Under this condition , It means that the class was attacked during the time when the poisoning took effect , So at this time, it is reasonable that our poisoning time can not simply add duration 了 , Because the attack time at this time overlaps with the last poisoning time , This part must be subtracted , What about this overlapping part , Namely timeSeries[i] - end, The result is negative , The effect of adding is to reduce .

---

Question no ：414, The third largest number

Subject portal , Poke poke began to solve the problem >>

---

class Solution {
    
    public int thirdMax(int[] nums) {
    
        if(nums.length == 1){
    
            return nums[0];
        }
        if(nums.length == 2){
    
            return (nums[0] > nums[1]) ? nums[0] : nums[1];
        }
        Arrays.sort(nums);
        int ret = 0;
        int count = 0;
        int tmp = Integer.MAX_VALUE;
        for(int i = nums.length - 1 ;i >= 0;i--){
    
            if(nums[i] != tmp){
    
                count++;
            }
            tmp = nums[i];
            if(count == 3){
    
                break;
            }

        }
        if(count == 3){
    
            return tmp;
        }else{
    
            return nums[nums.length - 1];// The array has no third largest number , for example  2 2 2 2 2 8;
        }
    }
}

Two special cases are listed separately , The array has only two numbers , Or the array has only one number , At this time, it is relatively simple to find the third largest number . Excluding these two cases , Sort the array first , So the array is arranged from small to large , Start traversing from the end of the array , As long as there is no quarrel tmp equal , Our count variable count++, Until we find the third largest number . Make use of count The reason for this is that the element has the possibility of repetition , So we sorted it out , The penultimate number is not necessarily the third largest number .

---

Question no ：628, The maximum product of three numbers

Subject portal , Poke poke began to solve the problem >>

---

class Solution {
    
    public int maximumProduct(int[] nums) {
    
        //  Three pointer traversal is too inefficient 
        Arrays.sort(nums);
        int n = nums.length - 1;
        return Math.max(nums[0]*nums[1]*nums[n],nums[n-2]*nums[n-1]*nums[n]);
    }
}

Sort the array first , There are three cases of maximum product ：
1, The last three numbers of the array are all positive numbers , The maximum product is the product of the last three numbers .
2, The last three numbers of the array are all negative numbers , The maximum product is also the product of the last three numbers . The size comparison of negative numbers is opposite to that of positive numbers
3, There are positive and negative numbers in the last three numbers of the array , It can only be the product of the smallest two negative numbers and the largest positive numbers .
So take it all into consideration , That is to integrate and use Math.max Just find the maximum , It includes the above situation .

---

Last , Today's article sharing is relatively simple , That's it , If you think it's good , Please help me to praise , Thank you very much. ！🥰🥰🥰