Jumping game II in question 45 of C language power deduction. Ergodic jump

Here's an array of nonnegative integers nums , You are first in the array .

Each element in the array represents the maximum length you can jump at that location .

Your goal is to reach the last position of the array with the least number of jumps .

Suppose you can always reach the last position of the array .

Example 1:

Input : nums = [2, 3, 1, 1, 4]
     Output : 2
     explain : The minimum number of jumps to the last position is 2.
     From the subscript for 0 Jump to subscript 1 The location of , jump 1 Step , Then jump 3 Step to the last position of the array .

     Example 2 :

     Input : nums = [2, 3, 0, 1, 4]
     Output : 2

     Tips :

    1 <= nums.length <= 104
    0 <= nums[i] <= 1000

1. If one acts as Take off point The distance that a grid can jump is 3, Then it means that 3 Each grid can be used as Take off point . Can be for everyone who can do Take off point All the squares try to jump once , hold Can jump to the farthest distance Constantly updated .
2. If we start from this Take off point The take-off is called the first 1 Time jumping , So from the back 3 A grid take-off all It can be called the first 2 Time jumping .
3. therefore , When a jumping At the end , Start with the next grid , Up to now Can jump to the farthest distance , all It's the next time jumping Of Take off point .

        31. For every time jumping use for Loop to simulate .
        32. After one jump , Update next time Take off point The scope of the .
        33. Jump in a new range , to update Can jump to the farthest distance .

Record jumping frequency , If you jump to the end , And you get the result .

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <string.h>


int jump(int* nums, int numsSize) 
{
	int end = 0, farthest = 0;
	int jumps = 0;
	int i = 0;
	for (i = 0; i < numsSize - 1; i++) 
	{
		// Judge the current farthest reachable position 
		farthest = fmax(nums[i] + i, farthest);
		// Update the boundary and increase the number of jumps  1.
		if (end == i) 
		{
			jumps++;
			end = farthest;
		}
	}
	// Return the number of jumps 
	return jumps;
}