[Function add]

1. Add leetcode solutions with dp tag.

Author: Seanforfun

leetcode/303. Range Sum Query - Immutable.md

@@ -111,4 +111,34 @@ class NumArray {
  * NumArray obj = new NumArray(nums);
  * int param_1 = obj.sumRange(i,j);
  */
-```
+```
+
+### Third time
+* Method 1: dp[n]: saves the sum of values up to index n(included).
+	* save the sum values.
+	* transfer function: dp[i, j] = dp[j] - dp[i] + nums[i]
+	```Java
+	class NumArray {
+		private int[] nums = null;
+		private int[] sum = null;
+		public NumArray(int[] nums) {
+			this.nums = nums;
+			int sum = 0;
+			this.sum = new int[nums.length];
+			for(int i = 0; i < nums.length; i++){
+				sum += nums[i];
+				this.sum[i] = sum;
+			}
+		}
+		
+		public int sumRange(int i, int j) {
+			return this.sum[j] - this.sum[i] + this.nums[i];
+		}
+	}
+
+	/**
+	 * Your NumArray object will be instantiated and called as such:
+	 * NumArray obj = new NumArray(nums);
+	 * int param_1 = obj.sumRange(i,j);
+	 */
+	```

leetcode/70. Climbing Stairs.md

@@ -81,3 +81,21 @@ class Solution {
         return dp[n];
     }
 }
+```
+
+### Third time:
+* Method 1: dp
+	* dp[n]: n is from 0 to number of stairs
+	* initial: dp[0] = 1, means there is one way to climb to stair 0.
+	* transfer function: dp[i] = dp[i - 1] + dp[i - 2].
+	```Java
+	class Solution {
+		public int climbStairs(int n) {
+			int[] dp = new int[n + 1];
+			dp[0] = 1;
+			for(int i = 1; i <= n; i++)
+				dp[i] = dp[i - 1] + (i >= 2 ? dp[i - 2]: 0);
+			return dp[n];
+		}
+	}
+	```

leetcode/746. Min Cost Climbing Stairs.md

@@ -0,0 +1,45 @@
+## 746. Min Cost Climbing Stairs
+
+### Question
+On a staircase, the i-th step has some non-negative cost cost[i] assigned (0 indexed).
+
+Once you pay the cost, you can either climb one or two steps. You need to find minimum cost to reach the top of the floor, and you can either start from the step with index 0, or the step with index 1.
+
+```
+Example 1:
+
+Input: cost = [10, 15, 20]
+Output: 15
+Explanation: Cheapest is start on cost[1], pay that cost and go to the top.
+
+Example 2:
+
+Input: cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
+Output: 6
+Explanation: Cheapest is start on cost[0], and only step on 1s, skipping cost[3].
+```
+
+Note:
+1. cost will have a length in the range [2, 1000].
+2. Every cost[i] will be an integer in the range [0, 999].
+
+### Solution
+* Method 1:dp
+	* dp[n + 1]: n + 1 means the top of the stairs whose cost is 0.
+	* Initialization: dp[0] = cost[0], dp[1] = cost[1]: we can either start from 0 or 1.
+	* transfer function: dp[i] = min(dp[i - 1], dp[i - 2]) + cost[i]
+	```Java
+	class Solution {
+		public int minCostClimbingStairs(int[] cost) {
+			int n = cost.length;
+			int[] dp = new int[n + 1];
+			dp[0] = cost[0];
+			dp[1] = cost[1];
+			for(int i = 2; i <= n; i++){
+				int cur = i == n ? 0: cost[i];
+				dp[i] = cur + Math.min(dp[i - 1], dp[i - 2]);
+			}
+			return dp[n];
+		}
+	}
+	```