[leetcode] Range Sum Query - Immutable

阿新 • • 發佈：2018-07-15

cti lin [] arr change interview nts 總結 fin

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Note:

You may assume that the array does not change.
There are many calls to sumRange

function.

分析：翻譯一下，給定一個數組nums，要求找（i，j）位置中間的和，也就是nums[i]+...+nums[j]。最差的方法當然是一個一個求和，但是因為這個題目很巧妙，他在一個題目中設計了很多用例，也就是說很多計算結果會被重復用到，因此需要不能單純的來一個用例就求和一次，要保存中間結果。很容易就想到用動態規劃來做數組的求和，維護一個dp數組，其中dp[i]表示從0到第i位置的和，狀態轉移方程也很容易就得到了：dp[i] = dp[i-1]+nums[i]，那麽求（i，j）的和就是dp[j]-dp[i-1]就好了。代碼如下：

 1 class NumArray {
 
 2     int[] dp;
 3     public NumArray(int[] nums) {
 4         if ( nums == null || nums.length == 0 ) return ;
 5         dp = new int[nums.length];
 6         dp[0]=nums[0];
 7         for ( int i = 1 ; i < nums.length ; i ++ ){
 8             dp[i] = dp[i-1]+nums[i];
 9         }
10     }
11 
12 
     public int sumRange(int i, int j) {
13         return i==0?dp[j]:dp[j]-dp[i-1];
14     }
15 }

這裏要註意第4行。運行時間125ms，擊敗96.34%。

感覺還是有優化的空間的，因為這種方法考慮太多的特殊情況，能不能想辦法將特殊情況也總結進去。看了一下solution，神奇的發現用一個長度為nums.length+1的數組就可以避免兩種特殊情況的討論了。代碼如下：

 1 private int[] sum;
 2 
 3 public NumArray(int[] nums) {
 4     sum = new int[nums.length + 1];
 5     for (int i = 0; i < nums.length; i++) {
 6         sum[i + 1] = sum[i] + nums[i];
 7     }
 8 }
 9 
10 public int sumRange(int i, int j) {
11     return sum[j + 1] - sum[i];
12 }

設計很巧妙，可以學習。

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