58 lines
1.9 KiB
C++
58 lines
1.9 KiB
C++
// Source : https://leetcode.com/problems/min-cost-climbing-stairs/
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// Author : Hao Chen
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// Date : 2019-02-04
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/*****************************************************************************************************
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*
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*
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* On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).
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*
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* Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to
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* reach the top of the floor, and you can either start from the step with index 0, or the step with
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* index 1.
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*
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* Example 1:
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*
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* Input: cost = [10, 15, 20]
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* Output: 15
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* Explanation: Cheapest is start on cost[1], pay that cost and go to the top.
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*
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* Example 2:
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*
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* Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
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* Output: 6
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* Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].
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*
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* Note:
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*
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* cost will have a length in the range [2, 1000].
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* Every cost[i] will be an integer in the range [0, 999].
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*
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******************************************************************************************************/
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class Solution {
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public:
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int minCostClimbingStairs(vector<int>& cost) {
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return minCostClimbingStairs02(cost);
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return minCostClimbingStairs01(cost);
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}
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int minCostClimbingStairs01(vector<int>& cost) {
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vector<int> dp(cost.size() , 0);
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dp[0] = cost[0];
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dp[1] = cost[1];
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for (int i=2; i<cost.size(); i++) {
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dp[i] = min( dp[i-1], dp[i-2] ) + cost[i];
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}
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return min(dp[dp.size()-1], dp[dp.size()-2]);
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}
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int minCostClimbingStairs02(vector<int>& cost) {
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int dp1 = cost[0], dp2 = cost[1];
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for (int i=2; i<cost.size(); i++) {
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int dp = min( dp1, dp2 ) + cost[i];
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dp1 = dp2;
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dp2 = dp;
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}
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return min (dp1, dp2);
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}
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};
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