LEETCODE 45. Jump Game II

题目大意

给出一个正整数数组，每个元素代表当前位置所能跳转的最大步数，求能到达数组最后的的最少跳转次数，实际上是44题的一个变种，44题只是判断能否到达数组最后，本题多加一个要记录最少跳转次数的要求。
例如
A = [2,3,1,1,4]
The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

解题思路

也是贪心策略，记录所能到达最大下标，并且在更新最大下标的时候更新最小步数，当最大下标大于等于数组最后的时候，直接返回这个最小步数

实现代码

超时

class Solution {public:    int jump(vector<int>& nums) {        const int arraySize = int(nums.size());        int timesArray[arraySize];        int rightBound = 1, leftBound = 0;        for (int i = 1; i < arraySize; i++) {            timesArray[i] = INT_MAX;        }        timesArray[0] = 0;        while (leftBound < arraySize) {            int newRightBound = rightBound;            for (int i = leftBound; i < rightBound && i < arraySize; i++) {                for (int j = 1; j <= nums[i]; j++) {                    if (i + j + 1 > newRightBound) {                        newRightBound = i + j + 1;                    }                    if (i + j < arraySize && timesArray[i] + 1 < timesArray[i + j]) {                        timesArray[i + j] = timesArray[i] + 1;                    }                    if (i + j >= arraySize && timesArray[i] + 1 < timesArray[arraySize - 1]) {                        timesArray[arraySize - 1] = timesArray[i] + 1;                    }                }            }            if (rightBound == newRightBound) {                return newRightBound >= arraySize - 1?timesArray[arraySize - 1]:0;            }            leftBound = rightBound;            rightBound = newRightBound;        }        return timesArray[arraySize - 1];    }};

改善

class Solution {public:    int jump(vector<int>& nums) {        int index = 0, rightBound = 0, maxInd = nums[0], minStep = 1;        while (index < nums.size() - 1 && index <= rightBound) {            for (int i = index; i <= rightBound; i++) {                maxInd = max(maxInd, i + nums[i]);            }            index = rightBound + 1;            if (maxInd > rightBound) {                if (maxInd >= nums.size() - 1) {                    return minStep;                }                minStep++;                rightBound = maxInd;            } else {                break;            }        }        return 0;    }};