Leetcode problem brushing plan -- monotonic sequence

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  subject ： Monotonous sequence

Original link ： Power button

  analysis ： 

This question only requires monotonous , So we Consider monotonic increasing and monotonic decreasing condition

Ideas ： 

We can go through Array traversal The way , use Intermediate variable to record whether it is monotonous , If two array elements satisfy monotonicity , Add one to this variable , So traverse to the end of the array .

If the whole array is monotonous , Then the value of its corresponding intermediate variable must also increase to a fixed value ： The length of the array -1

 Java Language version ：

class Solution {
    public boolean isMonotonic(int[] nums) {
        //  Initialize two intermediate variables , The two intermediate variables correspond to monotonic increasing and monotonic decreasing respectively 
        int flag1 = 0, flag2 = 0; 
        
        for (int i = 1; i < nums.length; ++i) {
            //  At present, when the last two numbers are equal, they also satisfy monotonicity 
            if (nums[i] >= nums[i - 1]) { 
                ++flag1; //  An intermediate variable that represents the incremental situation flag1 Add one 
            }

            if (nums[i] <= nums[i - 1]) {
                ++flag2; //  An intermediate variable representing a decreasing situation flag2 Add one 
            }
        }
        //  The loop is executed nums.length - 1 Time , And only execute the qualified... Once at a time if sentence 
        //  For example, if every execution is incremental if sentence , At the end of the cycle, the value of the intermediate variable is naturally equal to nums.length - 1,
        //  That is, the array is monotonically increasing , If it is monotonically decreasing, the same is true 
        if (flag1 == nums.length - 1 || flag2 == nums.length - 1) {
            return true; //  Satisfying monotone return true
        }
        return false;
    }
}

C Language version ：

bool isMonotonic(int* nums, int numsSize){
int flag1 = 0;// Initialize intermediate variable 
    int flag2 = 0;
    int k = numsSize - 1;
    for(int i = 1; i <= k; ++i){  // Cycle only k Time , That is, to make the value of the intermediate variable equal to k; Two of the loops if Statements can only execute the same... At a time 
        if(nums[i] >= nums[i-1]){ // For example, if each execution if It's all monotonically increasing , At the end of the loop, the value of the intermediate variable is naturally equal to k, That is, the array is single increment , return true
            ++flag1;                                                                 
        }
        if(nums[i] <= nums[i-1]){
            ++flag2;
        }
    }
    if(flag1 == k || flag2 == k){ 
        return true;             
    }
    return false;
}

Okay , That's all for today's topic , I'll see you on the next topic