55. Jump Game

Problem:

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index.

Example 1:

Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
             jump length is 0, which makes it impossible to reach the last index.

Solution:

ONE

See 45. Jump Game II. If the range does not expand at some point, we know it is stuck.

/**
 * @param {number[]} nums
 * @return {boolean}
 */
let canJump = function (nums) {
  for (let l = 0, r = 1; r < nums.length; ) {
    let rNext = r;
    for (let i = l; i < r; i++) {
      const rNextAtmp = i + nums[i] + 1;
      if (rNextAtmp > rNext) {
        rNext = rNextAtmp;
      }
    }
    if (rNext <= r) {
      return false;
    }
    l = r;
    r = rNext;
  }
  return true;
};

TWO

If we view it backward, and if the range of nums[n-2] covers nums[n-1], then we can safely make n-2 the new destination point, and so on.

If nums[0] can cover the last destination point, it is good.

/**
 * @param {number[]} nums
 * @return {boolean}
 */
let canJump = function (nums) {
  let des = nums.length - 1;
  for (let i = des - 1; i > 0; i--) {
    if (nums[i] + i >= des) {
      des = i;
    }
  }
  return nums[0] >= des;
};

☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆

☆: .｡. o(≧▽≦)o .｡.:☆☆: .｡. o(≧▽≦)o .｡.:☆