Leetcode - Jump Game Two

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和Jump Game幾乎相同的想法，他們是DP。關鍵是使用數組maxNumbers[k]儲存的地方k步驟的話。序號的最遠範圍，注陣maxNumbers[]它遞增。

class Solution {public:const int MAXVALUE = 1 << 30;int findMinStepToIndex(int maxNumbers[],int maxSteps,int index){if (index == 0)return 0;int left = 1;int right = maxSteps;while (left < right){int m = (left + right) / 2;if (maxNumbers[m] < index){left = m + 1;}else if (maxNumbers[m] > index){if (maxNumbers[m - 1] < index)return m;else if (maxNumbers[m - 1] == index)return m - 1;elseright = m - 1;}else{return m;}}return (right + left) / 2;}int jump(int A[], int n) {int* maxNumbers = new int[n];//mark the max number that steps i can walk.int maxIndex = 0;int maxNumber = 0;//the max index we can walk to.maxNumbers[0] = 0;for (int i = 1; i < n; i++){maxNumbers[i] = 0;}int maxSteps = 0;for (int i = 0; i < n - 1; i++){if (maxNumber < i + A[i]){int cMinStep = findMinStepToIndex(maxNumbers, maxSteps, i);maxNumbers[cMinStep + 1] = i + A[i]; maxNumber = i + A[i];maxSteps = cMinStep + 1;if (maxNumber >= n - 1)return maxSteps;}}}};

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Leetcode - Jump Game Two